








































































































































































































































































































































































































































































































































































































































































































































































/ 








•* y 























1 


• ( 

V * 
















:l -Il'II.CJi 

THE DOMAIN OF GEOMETRY 
























































... 




iiiiiiiiiiiiLiiiii;iiiiiiiHiiiiii!iiii!iHiO!iijiiiiiiiiin!!hiii!iiiiininiiiii!iii;iiiMiiiii!uiiiiiii|||ji^||m^ .... 


WENTWORTH-SMITH MATHEMATICAL SERIES 


AL SERIES 


iiiiiiiiiiiiiiiiiiiiiiiiimjiMi, 



ESSENTIALS OF 
PLANE AND SOLID GEOMETRY 


BY 


DAVID EUGENE SMITH 





GINN AND COMPANY 

BOSTON • NEW YORK • CHICAGO • LONDON 
ATLANTA • DALLAS • COLUMBUS • SAN FRANCISCO 








































COPYRIGHT, 1923, BY DAVID EUGENE SMITH 
ENTERED AT STATIONERS’ HALL 
ALL RIGHTS RESERVED 

623.10 





NOV 24 I9i;3 


* j 


VLfjt gitfienaum jgregg 

GINN AND COMPANY • PRO¬ 
PRIETORS • BOSTON • U.S.A. 


©C1A7CB002 

•^0 i 



PREFACE 


Demonstrative geometry is taught for the purpose of 
giving the student an insight into deductive reasoning, of 
allowing him to know what it means to prove a statement, 
of giving him the privilege of '' standing upon the vantage 
ground of truth,of cultivating his habits of independent 
investigation, of developing his own rules in applied mathe¬ 
matics, and of stimulating his appreciation of the beauties 
of the science. 

In some schools the course of study permits of doing 
this work thoroughly, while in other schools the pressure 
upon the curriculum is such as to allow less time than 
might profitably be used. On this account it is necessary 
to adjust a textbook so that it may permit of such fiexi- 
bility in its use as will adapt it to curricula of various kinds. 
To accomplish this purpose the propositions and corollaries 
have been limited to those that are actually necessary for 
the proof of subsequent statements or that are needed for 
a considerable number of important exercises. The lists of 
propositions prepared under the authority of the National 
Committee on Mathematical Requirements and of the 
College Entrance Examination Board have been followed 
as closely as the best principles of sequence and selection 
seem to warrant. The exercises have been carefully se¬ 
lected and have been made so numerous that any school 
may find abundant material for a long and thorough course, 
while another school may easily limit the course without 
destroying the sequence. 

PS V 


VI 


PREFACE 


In general, the fundamental theorems are given first, 
ordinarily followed by the fundamental constructions to 
which the theorems lead. In this way the great basal prop¬ 
ositions are so grouped as to command the special atten¬ 
tion which they deserve. Indeed, for a brief course in 
geometry the other propositions, including the numerical 
work and circle measurement, might be omitted or else 
referred to informally in the relatively few cases in which 
they are needed in subsequent proofs. 

Among other topics the Supplement contains a treatment 
of the practical mensuration of plane and solid figures along 
the lines recognized by the College Entrance Examination 
Board as furnishing valuable replacement material for some 
of the more formal work in Books VI-VIII. This feature 
satisfies a frequent demand for a modem type of training in 
spatial perception and supplements the logical presentation 
of the standard propositions. 

Among the special features of the work may be men¬ 
tioned the selection and arrangement of propositions, the 
simplicity of language and of proofs, the introduction to 
independent demonstration, the statements of the plan of 
proof, the applications, the improved t\T)ography, and the 
emphasis secured through the framing of the diagrams. 

My long and intimate association with my lamented col¬ 
league, George Wentworth, who, unfortunately, died before 
this book was undertaken, and the life-long infiuence of 
the sound principles established by his father, George A. 
Wentworth, have, I venture to hope, qualified me to write 
in the spirit which has made the mathematical textbooks 
bearing the Wentworth name of such inestimable service 
to more than one generation of teachers and students. 


DAVID EUGENE SMITH 


CONTENTS 


INTRODUCTION 

PAGE 

I. Common Terms Explained. 1 

II. Definitions. 7 

III. Demonstrative Geometry.13 

BOOK I. RECTILINEAR FIGURES 

I. Fundamental Theorems.21 

II. Fundamental Constructions.67 

III. Inequalities.81 

IV. Attacking Originals.93 

BOOK II. THE CIRCLE 

I. Fundamental Theorems.109 

II. Loci.139 

III. Fundamental Constructions . ..147 

BOOK III. PROPORTION AND SIMILARITY 

I. Fundamental Theorems.157 

II. Fundamental Constructions.185 

III. Numerical Relations.191 

PS vii 















Vlll 


CONTENTS 


BOOK IV. AREAS OF POLYGONS 

PAGE 

I. Fundamental Theorems.201 

II. Fundamental Constructions.219 

III. Supplementary Constructions.223 

BOOK V. REGULAR POLYGONS AND THE CIRCLE 

I. Fundamental Theorems.227 

II. Fundamental Constructions.237 

III. Circle Measurement .247 

IV. General Review.257 

• 

BOOK VI. LINES AND PLANES IN SPACE 

I. Lines and Planes.275 

II. Dihedral Angles .297 

III. Polyhedral Angles.307 

BOOK VII. POLYHEDRONS, CYLINDERS, AND CONES 

I. Prisms . . . .*.313 

11. Pyramids.333 

III. General Polyhedrons . 347 

IV. Cylinders.351 

V. Cones.359 

PS 

















CONTENTS 


IX 


BOOK VIII. THE SPHERE 

PAGE 

I. Fundamental Theorems.375 

II. Spherical Polygons.387 

III. Mensuration.399 

IV. General Review. 415 

SUPPLEMENT 

I. Incommensurable Cases. 425 

II. Polyhedrons. 429 

III. Spherical Triangles. 441 

IV. Practical Mensuration. 449 

V. Recreations.481 

VI. History of Geometry.485 

VII. Suggestions to Teachers.491 

VIII. Important Formulas.497 

INDEX.499 


PS 















SYMBOLS AND ABBREVIATIONS 


The following are the most important symbols used : 


+ 

plus 

Z 

angle 

- 

minus 

A 

triangle 

x» • 

times 

□ 

rectangle 

/, : 

divided by 

EJ 

parallelogram 

V 

square root of 

© 

circle 

8/~ 

V 

cube root of 

st. 

straight 

= 

is equal to, equals. 

rt. 

right 


is equivalent to 

A \ A '', 

A-prime, A-second, 

a- 

square of a 

A '", . . 

A-third, • • • 


cube of a 

Ai , A2 , 

A-one, A-two, 


and so on 

As , • • - 

A-three^ • • • 

> 

is greater than 

Ax. 

axiom 

< 

is less than 

Post. 

postulate 


therefore 

Const. 

construction 

—>► 

tends to 

Def. 

definition 

II 

parallel 

Cor. 

corollary 

X 

perpendicular 

Iden. 

identical 


Symbols of aggregation are used as explained in the text. 

There is no generally accepted symbol for is congruent 
to.’’ The sign = is commonly employed, the context tell¬ 
ing whether equality, equivalence, identity, or congruence 
is to be understood ; but teachers often use =, =, or = for 
congruence, and or for similarity. The symbol = is 
also used for identity, but is rarely needed in geometry. 

There is no generally accepted symbol for arc.” Some 
teachers recommend using AB for ”arc AB/' and this 
symbol has certain advantages. 


PLANE GEOMETRY 


INTRODUCTION 

1. Common Terms Explained 

1. Nature of Geometry. We are now about to begin 
another branch of mathematics, one not chiefly relating 
to numbers, although it uses numbers, and not primarily 
devoted to equations, although it uses them, but one that is 
concerned principally with the study of forms, such as tri¬ 
angles, parallelograms, and circles. Many facts that are 
stated in arithmetic and algebra are proved in geometry. 

2. Terms already Known. The student already has con¬ 
siderable familiarity with the terms that he will need to 
use. For example, he has a fairly good idea of such terms 
as straight line, curve, right angle, acute angle, triangle, 
square, and circle. In the case of certain of these terms 
it is unnecessary and even undesirable for the student to 
give the time and thought essential to the wording of a 
careful definition. 

3. Precise Definitions. In the case of other terms, how¬ 
ever, precise definitions are necessary, for the reason that 
we make use of such definitions in proving certain impor¬ 
tant statements to be studied later. 

Unless the student is specifically told that it is neces¬ 
sary to memorize a definition, it will be sufficient if he is 
able to use the terms correctly. 


2 


COMMON TERMS EXPLAINED 


INTROD. 


4. Surface, Line, Point. The solid here shown has six flat 
faces, each being a rectangle. It is called a rectangular solid. 
Statements and terms that should be 
considered most carefully, although infor¬ 
mally, are printed in italic type. 

Each of the six flat faces is part 
of the surface of the solid, and each 
is itself called a surface. 

If each of these flat faces is so 
smooth that when a straight ruler 
lies upon it in any position all points 
of the ruler touch the surface, the flat face is called a plane 
surface or simply a plane, 

A surface has length and breadth, hut no thickness. 

In all such cases, examples in the classroom should be noticed. 

In the above figure some of the faces meet in lines, and 
these lines are the edges of the solid. 

The way in which faces and lines are named will be 
understood from the statement that the faces AEFB and 
ABCD meet in the line AB, 

A line has length hut neither breadth nor thickness. 

We may represent a line by a mark, but a mark is really a very 
thin solid made of chalk, ink, or some other writing material. 

We commonly speak of solids, surfaces, and lines as 
magnitudes. 

In the above figure the lines BC and CD meet in the point 
C, a vertex of the solid, and one of the eight vertices, 

A point has position but not size, 

A point, a line, a surface, or a solid, or any combination 
of these, is called a geometric figure or simply a figure. 
Plane geometry considers figures of which all parts lie 
in one plane. 









§§4,5 


LINES 


3 


5. Lines. The figures AB and m here shown represent 
straight lines. When no misunderstanding is likely to arise, 
a straight line is called simply a line. 

Thus, we speak of the line AB and ^ ^ 

the line m, meaning straight lines. ' 

Lines and surfaces are supposed to extend indefinitely 
far unless the contrary is stated. If we wish to speak of 
part of a line limited by two points, 

we call it a line segment or simply i —£-- 

a segment. In this figure, PQ is a line 
segment, since it is a definite part of the unlimited line I, 
If we wish to speak of a line beginning at a certain point 
O and extending indefinitely, we call it 



a ray. In this figure a, 6, c are rays. 


When no misunderstanding is likely 
to arise, it is customary to use the word ^Mine’^ instead 
of''segment^’or ''ray.'^ 

A line of which no part is straight is called a curve line 

or simply a curve. The line AB here -^ 

shown is a curve line. 

Two straight-line segments that can be placed one upon 
another so that their end points coincide are said to be 
equal, ^ C ^ 

In this figure, AB = CD, as may be seen by meas¬ 
uring with compasses. By putting one point of 
the compasses at C and the other at D, and then, 
without changing the opening of the compasses, 
putting one point at A and the other at B, we can 
transfer CD to AB. D 

In the line I here shown, AC is the sum of AB and BC\ 


that is. 


B c 


AC=AB+BC, 


Also, BC is the difference between AC and AB\ that is, 
BC=AC-AB, 









4 


COMMON TERMS EXPLAINED 


INTROD. 



6. Angles. If two rays proceed from the same point, they 
form an angle. In this figure the rays OA and OB form 
the angle AOB, The vertex of this angle is 
0 and the arms or sides are OA and OB. 

When no misunderstanding is likely to 
arise, an angle may be named by the letter 
at the vertex or by a small letter within 
the angle, as in the cases of angles O and 
m here shown. If three letters are necessary, the middle 
one represents the vertex, as in the angle AOB above. 

The size of an angle depends upon the amount of turn¬ 
ing necessary to bring one arm to coincide with the other. 
Thus, taking these 
compasses, we see 
that the first angle is 
less than the second, 
and that the second 
is less than the third. 

We commonly measure angles in degrees, a right angle 
being 90°. In the above case the three figures are angles 
of 40°, 90°, and 120°, approximately. c 

In this figure, angle AOB is less than 

angle AOC, angle AOC is greater than - ^ 

angle AOB, angle AOC is the sum of ^ ^ 

angles AOB and BOC, and angle AOB is the difference 
between angle AOC and angle BOC; that is, 

ZAOB<ZAOC, 

ZAOOZAOB, 

ZAOC = ZAOBA-ZBOC, 
and ZAOB = ZAOC - ZBOC. 



Students are advised to provide themselves with compasses, a ruler, 
and a protractor for drawing figures. 












§§6-8 


RECTILINEAR FIGURES 


5 


7. Rectilinear Figure. A figure which lies wholly in one 
plane and which represents a surface that is bounded by 
segments of straight lines is called a 
plane rectilinear figure or simply a recti¬ 
linear figure. 

The segments are called the sides of 
the figure, and the adjacent sides meet in the vertices 
of the figure. The sum of all the sides is the perimeter 
of the figure. 

In modern geometry the bounding line is also considered as the 
* figure, and the perimeter as the total length of this line. In this book, 
unless the contrary is stated, only those figures will be considered in 
which each of the angles within the figure is less than two right angles. 

8. Triangle. A rectilinear figure of three sides is called 

a triangle. C 

A triangle is conveniently lettered / N. 
as here shown. The small letters rep- y 
resent the sides and correspond to the / 
large letters at the opposite vertices. c 

The side upon which a triangle or any other rectilinear 
figure is supposed to stand is considered as the base of 
the figure. 

The vertex opposite the base of a triangle is called the 
vertex of the triangle. Although a triangle has three ver¬ 
tices, it has only one that is called the vertex. 

In the above triangle: 

The three vertices are A, B, Cj and the three sides are 
designated as a, 6, c or as BC, CA, AB respectively. 

The vertex is C and the base is c. 

The perimeter is a -|- 6 + c, or BC+ CA-\~AB. 

The angles are BAQ CBA, ACB. 

The various types of triangles and other common rectilinear figures 
will be considered later, when the necessity arises. 





6 COMMON TERMS EXPLAINED introd. 

Exercises. Review of Common Terms 

Draw the following figureSy writing the name under each: 

1. Rectangle. 4. Rays. 7. Rectilinear figure. 

2. Solid. 5. Triangle. 8. Straight line. 

3. Curve. 6. Angle. 9. Line segment. 

Draw a figure representing each of the following: 

10. The sum of two line segments. 

11. The difference between two line segments. 

12. The sum of two angles; of three angles. 

13. The difference between two angles. 

14. A rectangular solid has how many edges ? how many 
faces ? how many vertices ? 

15. By counting the edges, faces, and vertices of a 
rectangular solid find the number to be added to the num¬ 
ber of edges to equal the sum of the faces and vertices. 

This law, which is useful in the study of crystals, holds for all 
ordinary forms of solids bounded by planes. The student may be 
interested to try it with a pyramid or any other convenient solid. 

16. Use a ruler to find out whether the top of your desk 
is approximately a plane as described in § 4. 

Of course, no such surface is exactly a perfect plane. 

17. Draw four angles, a, 6, c, d such that a <b<c<d. 

Consult the table of symbols and abbreviations when symbols are 
not clearly understood. 

18. Draw a curve of such shape that a straight line can 
cut it in four points and only four. 

19. Draw a figure showing the number of points in which 
one straight line can intersect another. 


§§ 9-11 


DEFINITIONS 


7 


11. Definitions 

9. Nature of Definitions. In §§10~22 we shall consider 
certain definitions which are so important that the student 
will find it convenient to memorize them, at least in sub¬ 
stance, because they are frequently needed in proving 
other statements. 

It should be understood that these definitions can be 
turned around; that is, if we say that certain conditions 
make a right angle, it follows that a right angle implies 
these conditions. In other words, 

A definition can he inverted. 

For example, if the organ of sight is called an eye, then an eye is 
the organ of sight. 

This is mentioned at the present time because the student will 
occasionally find it convenient to invert a definition. 

10. Equal Angles. If either of two angles can be placed 
on the other so that they coincide, the ^ 

two are called equal angles, _ — 

For example, these two angles are equal, all lines 
being supposed to be indefinitely long. The amount 
of turning necessary to make one angle is evidently 
the same as that necessary to make the other. 

In speaking of two figures that resemble each other it is often con¬ 
venient to use primes (q in lettering one of them. In the above case 
ZA'O'B' is read angle A-prime 0-prime S-prime." 

11. Bisector. A point, a line, or a plane that divides a geo¬ 
metric magnitude into two equal parts is called a bisector 
of the magnitude. 

For example, M, the midpoint of the line ^ 

AB, is a bisector of the line. Common sense 
will tell the student the meaning of such simple 
terms as midpoint. 

Similarly, we may have a bisector of an angle; 
for example, OM bisects the ZAOB here shown. 








8 


DEFINITIONS 


INTROD. 


12. Straight Angle. If the arms of an angle extend in 
opposite directions so as to be in one straight line, the 
angle is called a straight angle, x 

For example, both x and y in this figure are B - ^ 

straight angles, x being formed by turning the y 

arm OA halfway around the vertex O. 

A straight angle contains 180°; hence two straight angles contain 360®. 


13. Right Angle. Half of a straight angle is called a 
right angle. 

For example, x and y are evi¬ 
dently halves of the st. ZAOB and 
hence they are right angles; w, v, 
and z are also right angles. 

It follows from the definition that two right angles make a straight 
angle and that four right angles fill the space about a point. 


y/ff 




14. Perpendicular. If one line meets another so as to 
make a right angle with it, either of the two lines is said 
to be perpendicular to the other. 

In each of these figures, R is the 
vertex of a right angle ; hence in each 
figure, a is perpendicular to 6, and h is 
also perpendicular to a. 

The line a is called a perpendicular to 6, and 6 a perpendicular to a, 

A line that is perpendicular to a line segment and also bisects it is 
called a perpendicular bisector of the segment. 

The point R in each figure is called the foot of the perpendicular 
to h, or the foot of the perpendicular to a. 

The terms horizontal, vertical, oblique, and slanting, referring to 
lines, are used informally in geometry with the usual meaning with 
which the student is familiar. 



15. Square. A rectilinear figure of four equal sides and 
four right angles is called a square. 

This figure is too well known to require illustrating. 

The line joining opposite vertices of a square is called the diagonal, 
a term which we shall define later in connection with other figures. 








§§ 12-18 


ANGLES 


9 


16. Angles further classified. An angle is called 

an acute angle if it is less than a right angle; 
an obtuse angle if it is greater than a right angle ; 
a reflex angle if it is greater than a straight angle. 

Obtuse Angle Reflex Angle 

Acute and obtuse angles are called oblique angles, and 
each arm is said to be oblique to the other arm. 

If a wheel turns through more than 180°, each spoke turns through 
a reflex angle. If it turns through 360°, each spoke turns through a 
perigon, a term occasionally convenient. The wheel may, of course, 
turn through as many degrees as we please. If we speak of an Z O, 
however, we mean the Z O less than 180° unless the contrary is stated. 

17. Adjacent Angles. Two angles which have the same 
vertex and a common arm between them 
are called adjacent angles. 

For example, in this figure AAOB and BOC 
are adjacent angles. 

18. Angles classified by Sums. If the sum of two angles 
is a right angle, each is called the complement of the other, 
and the two angles are called complementary angles. 

If the sum of two angles is a straight angle, each is 
called the supplement of the other, and the two angles 
are called supplementary angles. 



Complementary Angles Supplementary Angles 


It may be assumed that if the sum of two adjacent angles 
is a straight angle, their exterior sides form a straight line. 




Acute Angle 









10 


DEFINITIONS 


INTROD. 


19. Triangles classified as to Sides. A triangle is called 
an isosceles triangle when two of its sides are equal; 
an equilateral triangle when all its sides are equal. 


Isosceles Equilateral 

The word ''equilateral” means equal-sided. It is applied to any 
figure having equal sides. 

An equilateral triangle is a special kind of isosceles triangle. 

An isosceles triangle is usually represented as resting on the side 
which is not equal to either of the other sides. This side is called the 
base, and the opposite vertex is called the vertex of the triangle. Ancient 
writers often spoke of the equal sides as the legs of the isosceles tri¬ 
angle, the word " isosceles ” meaning equal-legged. 

If no two sides of a triangle are equal, the triangle is called a scalene 
triangle, but the term is not commonly used. 

20. Triangles classified as to Angles. A triangle is called 

a right triangle when one angle is a right angle; 

an obtuse triangle when one angle is an obtuse angle; 

an acute triangle when all its angles are acute angles ; 

an equiangular triangle when all its angles are equal. 

tx A 

Obtuse Acute Equiangular 

In a right triangle the side opposite the right angle is 
called the hypotenuse. 

The other two sides of a right triangle are often called simply the 
sides when no confusion is likely to arise. 

Since ancient writers usually represented the hypotenuse as the 
base, the other two sides were called the legs of the right triangle. 



Right 








§§ 19-23 


TRIANGLE AND CIRCLE 


11 


21. Circle. A closed curve lying in a plane and such that 
all its points are equally distant from a fixed point in the 
plane is called a circle. 

When we draw a circle we sometimes say that 
we describe a circle. Either word, "draw” or 
"describe,” may be used in this sense. When a 
circle is drawn with the compasses we often say 
that we construct it. 

22. Terms relating to a Circle. The point 
in the plane from which all points on the circle are equally 
distant is called the center of the circle. 

A circle is commonly named by the letter at the center. In the 
above figure we may designate the circle as the O O. 

Any .one of the equal straight-line segments which 
extend from the center of a circle to the circle itself is 
called a radius (plural radii 

A straight line through the center and terminated at 
each end by the circle is called a diameter. 

It is evident that a diameter is equal in length to two radii. 

Any portion of a circle is called an arc. 

The length of the circle, that is, the distance around the 
space inclosed, is called the circumference. 

Formerly the term circle was used to mean the part of the plane 
inclosed, and the bounding line was then called the circumference. 

An arc that is half of a circle is called a semicircle. The 
length of a semicircle is called a semicircumference. 

An arc less than a semicircle is called a minor arc ; an 
arc greater than a semicircle is called a major arc. 

The word '' arc ’ ’ alone may be taken to mean a minor arc. 

23. Lines of Elementary Geometry. The straight line and 
the circle, or parts of such figures, are the only lines used 
in elementary geometry. 


Arc 



Circl® 


12 


DEFINITIONS 


INTROD. 


Exercises. Meaning of Terms 

1. Draw four right angles in different positions. 

All the drawings required on this page may be made freehand or 
by the aid of a ruler as the teacher may direct. At present the pur¬ 
pose is to fix in mind the meaning of the terms. 

2. Draw four lines in different positions and then draw 
three lines perpendicular to each of the four lines. 

3. Draw a horizontal line and a vertical line that inter¬ 
sect. What kind of angle is formed ? 

4. Draw four acute angles of different sizes. 

5. Draw an obtuse angle that is equal to the sum of a 
right angle and one of the acute angles of Ex. 4. 

6. Draw any acute angle and then draw its complement 
and its supplement. 

The protractor may be used advantageously in such cases. 

7. Draw three straight lines intersecting by twos. They 
may determine one point or how many points ? 

If the word '’determine” is not clearly understood, it should be 
considered in class. We say that in general three lines determine three 
points, meaning that this is the greatest number that they deter¬ 
mine, although in special cases, as the student should show, they may 
determine two points, one point, or no point. 

8. Through how many degrees does the minute hand 
of a clock turn in I hr. ? in 20 min. ? in 45 min. ? in hr. ? 

9. If a radius 3f in. long is used in drawing a circle, and 
if the circumference is times the diameter, find the 
circumference. 

10. If the supplement of Z a; is 4 a;, how many degrees 
are there in each angle ? 

11. If the complement of Zm is 3 m, how many degrees 
are there in each angle ? 


§§ 24,25 


DEMONSTRATIVE GEOMETRY 


13 


III. Demonstrative Geometry 

24. Need for Demonstrative Geometry. In looking at geo¬ 
metric figures we often find that we make mistakes if we 
judge by appearances. It is partly on this account that 
we need to demonstrate the truth of our judgments. 

For example, state which is the longer line, AB or XY, 
and estimate how many sixteenths of an inch longer it is. 

P - 


Then test your results by measuring with the compasses 
or with a carefully marked piece of paper. 

Look at this figure and state whether AB and CD are 
both straight lines. If one of them 
is not a straight line, which one is 
it? Test your answer by using a 
ruler or the folded edge of a piece 
of paper. 


2:^ 





Look at this figure and state whether the line AB will, 
if prolonged, lie on CD. Test your 
answer by laying a ruler along the 
line AB, 

Look at this figure and state which of the three lower 
lines is AB prolonged. Then test 
your answer by laying a ruler 
along AB. 

25. Bases for Proof. The proofs 
of geometry are based upon cer¬ 
tain assumptions known as ax¬ 
ioms and postulates. Since these assumptions do not depend 
merely upon the observation of figures, but upon common 
sense, they are universally accepted as the foundations 
upon which we may safely build our work. 











14 


DEMONSTRATIVE GEOMETRY 


INTROD. 


26. Axiom. A general statement admitted without proof 
is called an axiom. The following axioms should be memo¬ 
rized ; others will be assumed when needed. 

All numbers and magnitudes referred to in the axioms are con¬ 
sidered as positive. 


1. If equals are added to equals, the sums are equal. 


For example, since 
and 

we see at once that 


9 + 


9 = 5-h4 
5 = 3-h2 
5 = 54-4-1-3-1-2 


or 


14 = 14 


Likewise, if a = 3 and 6 = 7, then a-}-6 = 3-|-7 = 10. 


2. If equals are subtracted from equals, the remainders 
are equal. 


For example, since 
and 

we see at once that 
or 


9 = 5-h4 

_3 = 24-1 

9-3 = 54-4-2-1 
6 = 6 


Likewise, if a = 10 and x = 3, then a — a; = 10 — 3 = 7. 


3. If equals are multiplied by equals, the products are 
equal. 


For example, since 
and 

we see at once that 
that is, 
or 


12 = 15-3 

_ 2 = 2 _ 

2 xl2 = 2 X 15-2 X 3 
24 = 30 - 6 
24 = 24 


Likewise, if \x=l, then a; = 2 x 7 = 14. 

4. If equals are divided by equals, the quotients are equal. 
For example, since 16 = 94-7 
we see at once that 16-^4 = (9-|-7)-j-4 
that is, 4 = I -f I 

4 = = 4 

The divisor must never he zero, division by zero having no meaning. 


or 







§26 


AXIOMS 


15 


5. A number or magnitude may he substituted for its 
equal. 

For example, if a + a; = 6 and \ix = y, then a-\-y = h. 

If 6> £c and \i x = y, then h>y. 

x = h — a and \i y = h — a, then x = y. 

The student should make up other examples to illustrate this axiom. 

As a special case this axiom is often stated as follows: Quantities 
equal to the same quantity are equal to each other. 

The word '' quantity ’ ’ here refers to numbers or magnitudes. 

6. Like powers or like roots of equal numbers are equal. 

That is, if a: = 2, then x^ = 2^, or x^ = 4. Also, if x^ = 27, then a; = 3. 

7. If equals are added to or subtracted from unequals, or 
if unequals are multiplied or divided by equals, the results 
are unequal in the same order. 

This means that \i x>y and if a = 6, then 


x-\-a>y -\-h 
X — a>y — h 


ax > by 
X -i- a> y -i-b 


The student should illustrate each of the above cases by numerical 
examples, using the values x = 10, y = 5, a = 6 = 2, or others if desired. 

If X <y the above signs of inequality will all be reversed. 

8. If unequals are added to unequals in the same order, 
the sums are unequal in the same order; if unequals are 
subtracted from equals the remainders are unequal in 
reverse order. 

If a >b, c> d, and x — y, then a + c > b-\-d, and x — a <y — b. 

The student should illustrate as in Ax. 7. 

9. If the first of three quantities is greater than the 
second, and the second is greater than the third, then the 
first is greater than the third. 

Thus, if a> 6 and if 6 > c, then a>c. 

10. The whole is greater than any of its parts and is equal 
to the sum of all its parts. 


16 


DEMONSTRATIVE GEOMETRY 


INTROD. 


27. Postulate. In geometry a geometric statement ad¬ 
mitted without proof is called a postulate. The following 
postulates of plane geometry should be memorized; others 
will be assumed when needed. 

In considering the postulates, the student should draw a figure to 
illustrate each one. 

1. One straight line and only one can he drawn through 
two distinct points. 

This postulate is sometimes more conveniently expressed 
in one of the following forms: 

Two distinct points determine a line. 

Two straight lines cannot intersect in more than one point. 

For if they intersected in two points, the lines would coincide. 

Post. 1 may be given as the authority for any one of the above 
three statements. 

2. A straight-line segment can he produced to any 
required length. 

To produce AB is to extend it through B\ ^ _ 

to produce BA is to extend it through A. ^ 

In the figures in this book, lines produced are generally represented 
by dotted lines, as shown in § 48. 

3. A straight-line segment is the shortest path between 
two points. 

Since distance in a plane is measured on a straight line, this postulate 
is sometimes stated as follows: A straight line is the shortest distance 
between two points. More properly speaking, however, distance is the 
length of the line instead of the line itself. 

4. In a plane one and only one circle can he constructed 
with any given point as center and any given line segment 
as radius. 

From the definition of a circle and from this postulate we see and 
may hereafter state that all radii of the same circle are equal. 



§27 POSTULATES 17 

5. Any figure can he moved without altering its shape 
or size. 

That is, we may think of a triangle as moved about without any 
change in shape or size, and similarly for any other figure. 

6 . All straight angles are equal and all right angles are 
equal. 

The second part of the postulate follows from the first, because a 
right angle is half a straight angle. 

7. A line segment can he bisected, and in one and only 
one point. 

The student should show the reasonableness of this postulate by 
means of a figure. 

8 . An angle can he bisected, and by one and only one line, 

9. Angles which have equal complements or equal supple¬ 
ments are equal. 

For example, if the complement of Z a; is 22°, and the complement 
of Z 2 / is also 22°, this means that 

a;+ 22° = 90° 

and 2 /+ 22° = 90°. 

Then cc + 22° = 2/ + 22°. Ax. 5 

.\x = y. Ax. 2 

10. There is one and only one line which, passing through 
a given point, is perpendicular to a given line. 

Since a perpendicular to a line makes a right angle with it, and 
since we cannot, in the first of these 
figures, have ZAOB =ZAOC (Ax. 10), 
we cannot have two perpendiculars 
through O. 

If a line swings about P as a center, 
it may be assumed for the moment that 
there is only one position at which PQ is J- to Z. It is easily proved 
later that this assumption is true. 

As the student proceeds he will find that some of the other postu¬ 
lates, assumed for the present as true, can also be proved. 







18 DEMONSTRATIVE GEOMETRY introd. 

28. Theorem. A statement which is to be proved is called 
a theorem. 

For example, it is stated in arithmetic that the square of the hypote¬ 
nuse of a right triangle is equal to the sum of the squares of the other 
two sides. This statement is one of the most important theorems of 
plane geometry, and we shall prove it later. 

29. Problem. A construction which is to be made so that 
it shall satisfy certain given conditions is called a problem. 

For example, it may be required to construct an angle equal to a 
given angle. This construction will be made in § 106. 

30. Proposition. The statement of a theorem to be proved 
or of a problem to be solved is called a proposition. 

In geometry, therefore, a proposition is either a theorem or a 
problem. We shall find that the first group of propositions is made 
up of theorems. After we have proved a number of theorems we shall 
solve some of the most important problems. 

31. Corollary. A statement that follows from another 
statement with little or no proof is called a corollary. 

For example, since we admit that all straight angles are equal, it 
follows as a corollary that all right angles are equal, since a right 
angle is half a straight angle. 

32. How Propositions are Proved. We have said that we 
are now about to prove our statements in geometry, and 
we shall first see what is meant by a proof. For this pur¬ 
pose we shall take a simple proposition concerning vertical 
angles, a term which we must first define. 

33. Vertical Angles. When two angles have the same 
vertex and the sides of one are prolon¬ 
gations of the sides of the other, these 
angles are called vertical angles. 

In the figure here shown, x and z are vertical 
angles, and so are w and y. 




§§ 28-34 


HOW PROPOSITIONS ARE PROVED 


19 


34. Study of a Figure. Suppose that we consider the 
question of vertical angles with respect to the figure here 
shown. Does there appear to be in the 
figure any other angle equal to a; ? If 
so, which angle is it? 

The amount of turning of the ray 
OA about O to make the Za; is the 
same as the amount of turning of what other ray about O 
to make the Z 2 ; ? 

Then how does the amount of turning necessary to 
produce any angle compare with the amount of turning 
necessary to produce its vertical angle ? 

What does this lead you to infer as to the equality of 
X and zl as to the equality of any other vertical angles ? 

Let us now see how we can prove that any angle is equal 
to its vertical angle by referring to the axioms or postulates 
instead of considering the amount of turning necessary to 
produce the two angles. 

In the above figure, which angle is the supplement of 
both X and z ? 

Then how does the supplement of x compare with the 
supplement of zl 

What does Post. 9 tell us with respect to angles which 
have equal supplements? 

What can then be said about the equality of x and z ? 

What other two angles in the figure are equal ? 

Write and complete the following statement: 

If two lines intersect, the vertical • • •. 

The student has now seen how to prove a proposition, not by trusting 
to appearances but by depending only upon a definition and a postu¬ 
late. The definition was that of the supplement of an angle (§ 18), and 
the postulate was Post. 9 as mentioned above. In § 35 we shall show 
how this proof looks when stated more systematically and in proper 
geometric form. 



20 


DEMONSTRATIVE GEOMETRY 


INTROD. 


Specimen Proposition. Vertical Angles 

35. Theorem. If two lines inter sect, the vertical 
angles are equal. 



Given the lines AC and BD intersecting at O and forming 
A X, y, and z as shown. 

Prove tho,t x — z. 

Proof. a; + 2 / = ast.Z, §12 

because their arms extend in opposite directions so as to 
be in one st. line. 

Likewise, y + z = Sist.Z.. § 12 

2 / is the supplement of x and also of z. § 18 
If the sum of two A is a st. A, each is called the supplement 
of the other. 

.\X = Zy Post. 9 

because A which have equal complements or equal 
supplements are equal. 


36. Nature of a Proof. From § 35 it is seen that there are 
three steps in proving a theorem: (1) stating what is given 
(sometimes called the hypothesis), (2) stating what is to be 
proved (sometimes called the conclusion), and (3) giving the 
proof, each statement of which is supported by a definition, 
an axiom, a postulate, or a proposition previously proved. 












BOOK I 


RECTILINEAR FIGURES 


I. Fundamental Theorems 


37. Congruent Figures. If two figures have exactly the 
same shape and size, they are called congruent figures. 

For example, the two triangles shown below (§ 38) are congruent 
(con'gru-ent) figures, and are said to be congruent. Similarly, two 
circles with equal radii are congruent. 

If two figures can be made to coincide in all their parts, 
they are congruent figures. 

By the parts of a figure we mean the sides, angles,' and surface. 

38. Corresponding Parts. It is customary to letter the 

angles of a triangle by capitals arranged about the figure 
in counterclockwise order; that is, reading about the figure 
in the direction oppo- c c‘ 



site to that in which 
the hands of a clock 
move. 


Exceptions to this custom are mentioned later, as occasion arises. 

In the triangles shown above, A' corresponds to A, R' cor¬ 
responds to B,C' corresponds to C, a' corresponds to a, and 
so on; that is, these pairs of parts are respectively equal. 
It IS therefore evident that 

In two congruent figures the parts of one figure are equal 
respectively to the corresponding parts of the other figure. 

Some writers speak of corresponding parts as homologous parts. 


21 




22 


FUNDAMENTAL THEOREMS 


BOOK I 


39. Inference as to Congruent Triangles. When we examine 
two triangles we may easily infer certain facts relating to 
them. For example, as we look at these triangles, in which 


ZA = ZA', b = b', 
c = c', the triangles s 
to be congruent, 
question is: Are 



necessarily congruent? 

It aids the eye if we mark the equal corresponding parts 
in some such way as the one used in the above figures. 

In order to aid the beginner, in the figures of Book I the important 
lines used in the proofs are made heavier than the others and the 
important angles are appropriately marked. This scheme is not used 
after Book I. 

Teachers will see the objections to the use of colored crayons to 
designate corresponding parts except, perhaps, in the case of a few 
propositions. The student should early become familiar with the tools 
that he will actually use, the black lead pencil and the white crayon. 

To prove that the two triangles are congruent let us see 
if one triangle can be placed upon the other so as to coin¬ 
cide with it. To help us see this clearly we may, if we wish, 
cut two triangles out of paper. 

Suppose that A ABC is placed upon AA'B'C' so that the 
point A lies on the point A' and c lies along c'; then where 
does the point B lie, and why ? 

On what line does b then lie, and why ? 

Then where must C lie, and why ? 

Having found where B and C lie, where does a lie ? 

What have we now shown with respect to the coinciding 
of A ABC with AA'B'C'l Are the triangles congruent? 

Complete the following statement: If two sides and the 
included angle of one triangle are equal respectively to two 
sides and the included angle of another, the triangles • • •. 

The statement and formal proof are given in § 40. 




§§ 39. 40 


FIRST CONGRUENCE THEOREM 


23 


Proposition 1. Two Sides and Included Angle 

40. Theorem. If two sides and the included angle of 
one triangle are equal respectively to two sides and the 
included angle of another, the triangles are congruent. 



Given the k^ABC and A'B'C with c = c', b = h\ and 
ZA = ZA'. 

Prove that A ABC is congruent to AA'B'C'. 

The plan is to place one upon the other and show that they coincide. 

Proof. Place I\ABC upon AA'B'C' so that A lies on A' and 
c lies along c\ C and C' lying on the same side of c'. Post. 5 


Then 

B lies on B\ 

because c is given equal to c'; 



b lies along h', 

because ZAis given equal to ZA'; 


and 

C lies on C', 

because b is given equal to b'. 


Hence 

a coincides with a'. 

Post. 1 


One st line and only one can be drawn through two distinct points. 

A ABC is congruent to AA'B'C'y §37 

by the definition of congruent figures. 

This method of proof is called the method of superposition. 

PS 









24 


FUNDAMENTAL THEOREMS 


BOOK I 


Exercises. First Congruence Theorem 


1. If ABCD is a square and P is the midpoint 
prove that PD =PC. 

The student should write the work in the following 
form: 

Given a square ABCD and P, the midpoint of AB. 

Prove that PD = PC. 

Proof. AP = BP, ^ 

because P is given as the midpoint of AB. 

AD = BC. 

(Give the reason from ^15.) 

ZA=ZB. 

(Give the reasons from ^15 and Post. 6.) 

Hence • • • (state what follows from § 40 and give the reason). 

.\PD = PC. 

(Give the statement at the end of § 38.) 



When proofs are written on wide sheets of paper, some teachers 
require students to rule the page vertically in the center and to write 
the statements on the left side of the line and the full authority for 
each statement on the right side. Such an arrangement is sometimes 
convenient, although it is not as concise as the form suggested above, 
which is used in many standard textbooks. 


2. In this figure, if Z.A=Z.B, if M bisects and if 
AY=BX, prove that MY=MX. 

The student should begin the work as follows: 

Given ZA = ZB, M bisecting AB, and AY = BX. 

Prove that MY = MX. 

In the proof the student should see that he can show that MY = MX 
if he can show that /\AMY is congruent to L.BMX, and that he can 
show this if • • •, and so on. 

When two figures are arranged as above, with the corresponding 
letters of one in an opposite order from those of the other, it is much 
better to read^one set counterclockwise and the other clockwise, as in 
the above statement, so as to have the letters correspond more clearly. 






40 


FIRST CONGRUENCE THEOREM 


25 



3. In the square ABCD the points P, Q, P, S bisect the 
consecutive sides. Prove that PQ = QR = rs = SP, 

In this case the student will save time by first 
proving that PQ = QR, beginning as follows: 

Given the square ABCD with P, p, P, S bisecting 
ABj BC, CD, DA respectively. 

Prove that PQ = QR. 

In the proof show first that 

' AB=BC = CD; 

then that PB = BQ = QC = CR; 

then that ZB = ZC. 

Then show that APBQ and QCR are congruent. 

What follows ? . 

It is now unnecessary to prove the other triangles congruent, 
fer evidently this can be done in precisely the same way. Simply 
write, '' Similarly, the other A are congruent, and hence PQ = QR = 
RS=SP.’* When such methods of shortening the proof are used, the 
student must be sure that the cases are exactly similar. 

4. Prove that to determine the distance AB across a 
pond one may sight from A across a 
post P, place a stake at A’ making 
PA'=AP, then sight along BP making 
PB'—BP, and finally measure A'B'. 

What is given? What is to be proved? Write 
these statements and then write the proof. 

5. Show how to find the distance from a point P west 
of a hill to a point Q east of the hill, 

using the figure here shown. 

State what measurements you would make 

on the ground. Then write the proof as in the -——^P^ 

preceding cases. 

In all such cases of outdoor measurement the land on which the 
triangles are laid out is supposed to be a horizontal plane unless the 
contrary is stated. 








26 


FUNDAMENTAL THEOREMS 


BOOK I 


6. In the square ABCD here shown, prove that AC = BD, 

Begin as follows: 

Given the square ABCD. 

Prove that AC = BD. 

The student should attack such an exercise by 
saying to himself, I can prove this if I can prove 
that; I can prove that if I can prove this third 
statement,” and so on until he finds something already proved. He 
should then reverse this process, beginning with a proposition already 
proved and ending with the statement to be proved. 

7. In this figure, AD=BC and each is 
J_ to AB. What do you infer as to the 
relation of AC to BDl Prove the cor¬ 
rectness of your inference. 

8. If ABCD is a square, if P bisects CD, and if BM is 
made equal to AN, as shown in the figure, rt , P . n 
prove that PM = PN. 

"I can prove this if I can prove that A- 

and-are congruent. I can prove that these 

angles are congruent if • • •.” 

9. In this figure, AD = BC, each is JL 
to AB, and DP = CQ. What do you infer 
as to the relation of A APB to Z.BQA and 
of PB to QA ? Prove the correctness of 
your inferences. 

10. Suppose that it is known that a machine will work if 
three certain wheels properly gear into three other wheels. 
Suppose also that it is given that wheel a gears into wheel 
a\ that it can be shown that wheel h gears into wheel h\ 
and that it can then be shown that wheel c gears into 
wheel c\ What follows? 

An occasional exercise like Ex. 10 may be discussed for the sake of 
training in transferring geometric reasoning to other lines. 













§41 


ISOSCELES TRIANGLE 


27 


41. Inference as to an Isosceles Triangle. If we examine 
the isosceles triangle here shown, we can make several 
inferences; among them, that if 
6 = c, 

then Z5 = ZC. 

We have proved one proposition about 
equal angles (§ 35), but since that re¬ 
ferred to vertical angles it does not help us in this case. 

We have also proved,a proposition about congruent tri¬ 
angles (§40), and congruent triangles have equal angles. 
Possibly we may be able to prove that Z5=ZC if we can 
divide A ABC into two congruent triangles. 

In order to use § 40 we must have two sides and the 
included angle of one triangle equal respectively to two 
sides and the included angle of another triangle; hence in 
order to get two equal angles let us suppose that AM is 
the bisector of ZA (Post. 8). 

Dotted lines are used to represent such auxiliary lines as AM, which 
are inserted to assist in a proof. In speaking of ZA we mean the 
ZB AC, the original angle at A, and so in all similar cases. 

Then in A ABM and ACM, what is the relation of c to 6 ? 

What is the relation oixtoy with respect to size ? Why ? 

What line is the same in A ABM and ACM; that is, what 
line is common to the two triangles ? 

Then what parts of one triangle have you shown to be 
equal to what parts of the other triangle ? 

What can you say as to the congruence of the triangles, 
and what is the authority for the statement ? 

What can you say as to the relation of ZB to ZC? 

Complete the following statement: 

In an isosceles triangle the angles opposite the equal • • •. 

The statement and formal proof are given in § 42. 





28 


FUNDAMENTAL THEOREMS 


BOOK I 


Proposition 2. Isosceles Triangle 

42. Theorem. In an isosceles triangle the angles oppo¬ 
site the equal sides are equal. 


C 


Given the isosceles A ABC with b = c. 

Prove that /iB = /LC. 

The plan is to prove two A congruent. 

Proof. Let p be the bisector of Z A, meeting BC at M. 
Then in A ABM and ACM it is given that 

c = b. 

Further, x = y, § 11 

because p bisects ZA; 

and side p is common to both A. 

.*. A ABM is congruent to A ACM. § 40 

If two sides and the included Z of one A are equal respectively to 
two sides and the included Z. of another, the A are congruent. 

.\ZLB=z:c, §38 

because they are corresponding parts of congruent figures. 

43. Corollary. An equilateral triangle is 
equiangular. 

Because 6 = c(why?), what follows as to ZR 
and ZC? Why? Now prove that ZA = ZR 
Why does ZA = ZC? Write out the full proof. ^ 




a 


C 










ISOSCELES TRIANGLE 


29 


Exercises. Isosceles Triangles 

1. In this figure, which represents the cross section of 
the attic of a house, it is known that the rafters AB and 
AC are equal in length. Suppose that 
we find by measuring that ZB = 32°, 
but that we cannot conveniently pass ^ 
the partition p so as to measure ZC. 

If we are told that ZC = 30°, is the information correct ? If 
not, what should it be? Upon what proposition does the 
answer depend ? 

2. This figure represents a square ABCD 
separated, into two triangles by the diagonal 
AC, Which angles are equal by § 42 ? 

3. In the same figure state which triangles 
are congruent by § 40, and hence show what other angles 
are equal besides those found in Ex. 2. 

4. In this figure BA = BC and Z.DBA=/lDBC. Prove 

that /\ACD is isosceles. D 

We can prove that DA = DC if we can prove 
that AABD and CBD are congruent. We can 
prove this if we can show that § 40 applies. 

5. In Ex. 4 prove that DB is ± to AC, 

What two angles must be proved equal? In order to prove them 
equal, what two triangles must be proved congruent? 

6. In this figure PB = PC and ZAPB = ZAPC, Prove 

that A ABC is isosceles. a 

7. In the figure of Ex. 6 make a list 
of all the pairs of equal angles and 
prove each statement. 

The teacher will find it helpful to introduce such exploring exercises 
in connection with various other figures, letting the student discover 
for himself as many relations as possible. 











30 


FUNDAMENTAL THEOREMS 


BOOK I 


Proposition 3. Two Angles and Included Side 

44. Theorem. If two angles and the included side of 
one triangle are equal respectively to two angles and the 
included side of another^ the triangles are congruent 



Given the AARC and A'B'C' with ZA = ZA^ ZC= ZC', 
and &= &^ 

Prove that AABC is congruent to AA'R'C'. 

The plan is to place one upon the other and show that they coincide. 

Proof. Place A ARC upon A A'R'C' so that A lies on A' and 
h lies along h\ B and B' lying on the same side of b'. Post. 5 

Then C lies on C', 

because h is given equal to h'; 
c lies along c\ 

because /LA is given equal to /LA'; 

and a lies along a\ 

because /LC is given equal to Z. C'. 

Since R is on a and c, it lies on both a' and c\ and so 
lies on B\ the point common to both a' and c\ Post. 1 

Two st. lines cannot intersect in more than one point. 

. *. A ARC is congruent to A A'R'C', § 37 

by the definition of congruent figures. 










§44 


SECOND CONGRUENCE THEOREM 


31 



Exercises. Second Congruence Theorem 

1. In this figure, ABCD is a square, M is the midpoint 
of ABj and the lines MX and MY make 

equal angles with AB. Prove that A AMY is 
congruent to A BMX, What other angles in 
these triangles are equal, and why ? 

2. In the figure of Ex. 1, what angles of 
the figureMXCDY are equal, and why? 

3. In this figure, ABCD is a square and 
p = q. What other angles in the two tri¬ 
angles are equal? What lines are equal? 

Give the necessary proofs. 

4. Wishing to measure the distance across a river, some 
boys sighted from A to a point P. They 

then laid off the line AB at right 
angles to AP. They placed a stake at 
O, halfway from A to P, and laid off 
a perpendicular to AB at P, placing a 
stake at C on this perpendicular in 
line with O and P. They then found 
the width by measuring PC. Prove that they were right. 

5. In this figure, ZPCP = ZCPA, 

Z CBD=ZDAC, and PC = AD, Find the 
other equal lines and equal angles and 
prove that they are equal. ^ 

6. Wishing to find the distance PX, some boys measured 
ZXAB and ZABX with the aid of a protractor. They 
then made ZX'AB = ZXAB and 
ZABX' = ZABX, thus laying off the 
AABX', How could they then find ^ 
the distance BX ? On what proposition 
does this depend? 













32 


FUNDAMENTAL THEOREMS 


BOOK I 


45. Another Inference. Suppose that these two triangles 
have the three sides of one equal respectively to the three 
sides of the other; that is, sup- ^ 

pose that 

a = a\ 
h = h\ 

and c = c\ 

From the appearance of the 
triangles, what do you infer as 
to their congruence ? Would you draw the same inference 
if the three angles of one were equal respectively to the 
three angles of the other ? Draw figures to illustrate your 
answer to this second question. 



46. Examination of the Inference. In the case in which 
the three sides of one are equal respectively to the three 
sides of the other, see if you can give a satisfactory proof 
by placing A ARC upon AA'R'C', as in §§ 40 and 44. If not, 
try placing them as here shown, and 
drawing CC'. 

Because h = h\ what kind of trian¬ 
gle is AAC'C? Therefore what two 
angles of AAC'C are equal? 

Because a = a\ what kind of tri¬ 
angle is ARCC' ? Therefore what two 
angles of ARCC' are equal? 

By adding two pairs of equal angles, what can now be 
said as to the equality of ZC and ZC'? 

Can you now prove that A ARC and A'R'C' are congruent 
by using § 40 ? Try it. 

Complete the following statement: 

If the three sides of one triangle are .... 

The statement and formal proof are given in § 47. 


R 







§§ 45-47 


THIRD CONGRUENCE THEOREM 


33 


Proposition 4. Three Sides 


47. Theorem. If the three sides of one triangle are 
equal respectively to the three sides of another^ the 
triangles are congruent. 



Prove that A ABC is congruent to AA'B'C \ 

The plan is to adapt the figure to § 40. 

Proof. Place A ABC so that A lies on A\ c lies along c\ 


and C and C' lie on opposite sides of A’B Post. 5 
Then .R lies on .S', 

because c is given equal to c'. 

Drawing CC, we have b = b’, Given 

and hence ACC'A' =Z.C'CA'. §42 

In an isosceles A the A opposite the equal sides are equal. 

Also, since a = a', Given 

we have AlB'C'C= a1B'CC'. § 42 

Adding, a1CC'A'-\-a1B'C'C=a 1C'CA' -\-AlB'CC'; Ax. 1 
that is, ZB'C'A' =ZB'CA' (ZBCA). 

AA5C is congruent to AA'.R'C'. §40 

{State the theorem of § W as the reason.) 











34 


FUNDAMENTAL THEOREMS 


BOOK I 


Exercises. Third Congruence Theorem 

1. By placing three rods of different lengths end to end 
so as to form a triangle, can you form triangles of different 
shapes and sizes ? State the reason for your answer. 

2. Three iron rods are hinged at their ends as shown 
in this figure. Is the figure thus formed 
rigid; that is, can its shape be changed? 

State the reason. 

This explains the statement that a triangle is 
determined by its three sides. It also explains why 
the triangle is called a unit of rigidity in bridge building and in steel 
construction generally. 

3. Four iron rods are hinged at their ends as shown 
in this figure. Is the figure thus formed 
rigid ? If not, state two ways in which, by 
the addition of a single rod in each case, it 
can be made rigid. Upon what theorem 
does this depend? 

4. Draw a rough figure of the framework of a bicycle. 
State the reason or reasons for its rigidity. 

5. The following method is sometimes used for bisect¬ 
ing an angle by the aid of a carpenter’s square: Place the 
square as here shown so that the edges 
shall pass through A and B, two points 
equidistant from 0 on the arms of the q. 
given ZAOB, and so that AP = BP. 

Then draw OP. Show that OP bisects 
ZAOB. 

6. If in an equilateral triangle a line is drawn from one 
vertex to the midpoint of the opposite side, prove that 
the triangles thus formed are congruent. 










§47 


CONGRUENT TRIANGLES 


35 


Exercises. Review 


1. In the /\ABC it is given that AC = 5C and that CM 
bisects ZC. Prove that CM bisects AB. 

Draw the figure and say: ''I can prove this if I can prove •••.” 
Always attack an exercise in this way unless you see the proof at once. 

2. In this figure it is given that AM bisects ZA and is 
also JL to BC, Prove that A, ABC is 
isosceles. 

'' I can prove this if I can prove •••. But 
I can prove that by § 44. ” Now reverse the 
reasoning and write out the proof. 

3. In the A ABC it is given that Z.A—ZB, that P bisects 
AB, and that ZNPA=ZMPB, Prove 
that AN = BM. 

I can prove this if I can prove that AAPN 
is congruent to ABPM. I can prove that be- ^ 
cause I know • • •. ” 



B 




4. In this figure it is given that ZA=ZA\ ZB=ZB\ 
and AB=A'B'. Find the other equal 
lines and equal angles and prove that ^ 
they are respectively equal. 

Remember that BCA' is one of the angles. 



5. Prove that a perpendicular to the bisector of an angle 
forms an isosceles triangle with the arms of the angle. 

6. In the A ABC it is given that ZA=ZB and that 

AP and BQ are so drawn that ZQBA = c 

ZPAB, Prove that .RQ =AP. 

I can prove this if I can prove that AARQ 
is congruent to A RAP. I can prove that because 
I know •••.’’ A* 

7. In the figure of Ex. 6 state the pairs of equal angles. 








36 


FUNDAMENTAL THEOREMS 


BOOK I 


8. In the square ABCD it is given that the point 

P bisects CD and that PQ and PR are so drawn that 
ZQPC = 50° and ZPPQ = 80°. Prove that p ^ c 

PQ = PR. 

If Z QPC= 50° and ZRPQ = 80°, express ZD PR 
in degrees. 

In the ADRP and CQP, what parts are respec¬ 
tively equal, and why ? 

9. Prove that the line from the vertex of an isosceles 
triangle to the midpoint of the base is perpendicular to 
the base. 



10. In this section of a support for a heavy 
tank are both cross braces necessary for rigid¬ 
ity ? State the reason. If either one is unnec¬ 
essary, state a reason for having it there. 



11. Two isosceles triangles of different heights are con¬ 
structed on the same base and on the same side of the 
base. Prove that the line through their vertices bisects 
the angles at the vertices. 


12. In Ex. 11 suppose that the two isosceles triangles 
are constructed on opposite sides of the base. 


13. In this figure a = a' and b = h'. 
Prove that ZLP=zlQ. 

Hereafter the words ''prove that” will 
usually be omitted in the exercises when it is 
obvious that a proof is required. 



14. If from any vertex of a square there are drawn line 
segments to the midpoints of the two sides not adjacent 
to the vertex, these line segments are equal. 

15. From the propositions already studied write a com¬ 
plete statement of the different conditions under which 
two triangles are congruent. 










§§ 48,49 


EXTERIOR ANGLE 


37 


48. Exterior Angle. The angle included by one side of a 
plane figure and an adjacent side produced is called an 
exterior angle of the figure. ^ 


For example, e is an exterior angle of this 
triangle, and A A and C are called the non- 
adjacent interior angles. 



-X 


49. Inference as to an Exterior Angle of a.Triangle. In the 
above figure, which seems the larger, eorZA? e or Z.Cl 

Would your inference be the same if the triangle were 
of a different shape ? Consider, for exam¬ 
ple, this figure. 

We have thus far found no way of 
proving one angle greater than another, 
but we have found five different ways of proving one angle 
equal to another, one in § 35, three in §§ 40, 44, and 47, and 
one in § 42. 

Consider this 'figure, supposing that M bisects BC, that 
AM is drawn and is then produced so that MP = AM, and 
that BP is then drawn. 

Can you prove that and CAM 

are congruent? If so, can you prove 
that ZPBM = ZACM1 

Then is ZXBOZPBMl By what ^ 
axiom is this true ? 

Then how is ZXBC related to ZC, and why? 

Can you bisect AB and proceed in a similar way to 
show that ZABY>ZA ? If so, is ZXBOZA ? 

The student has now reached the point where he may profitably 
read the model proofs without such assistance as is given above. 

The model "proofs should not be memorized, but the student should 
read the theorems and try to work out the proofs for himself before 
reading those given in the book. The complete statement of the authority 
for each step of the proof should always he given, particularly where 
the reference number alone is quoted. 









38 


FUNDAMENTAL THEOREMS 


BOOK I 


Proposition 5. Exterior Angle of a Triangle 

50. Theorem. An exterior angle of a triangle is greater 
than either nonadjacent interior angle. 



Given the exterior AXBC of the l\ABC, 

Prove that Z.XBC > AC and that AXBC > AA, 

The plan is first to prove that ZXBC >ZPBM, which is equal to ZC. 

Proof. Let M bisect BC. Post. 7 

Draw AM and produce it so that MP = AM. Posts. 1, 2 
Draw BP. Post. 1 

The line BP lies within AXBC, for otherwise AP would 


cut either AX or AC produced in two points, which is 


impossible. 


Post. 1 

Then 

^BMP = ZCMA, 

§35 


BM = CM, 

§11 

and 

MP was made equal to MA. 


Then 

ABPM is congruent to A CAM, 

§40 

and hence 

APBM = ZC. 

§38 

But 

ZXBOZPBM, 

Ax. 10 

and hence 

ZXBOZC. 

Ax. 5 


Similarly, AABY > A A, and hence AXBC > AA. 

Draw the figure and give the proof of this last statement 









§§ 50-54 


PARALLEL LINES 


39 


51. Parallel Lines. Lines which lie in the same plane and 
cannot meet however far they may be produced are called 
parallel lineSy or simply parallels. 

For example, AB and. CD are parallel lines. We may think of them 
as edges of a strip of ribbon. Since the stu- ^ 
dent is already familiar with such lines, further 

illustrations are not necessary. ---S 

It should be observed that in the above definition the words *'in 
the same plane ’’ are essential. 


52. Postulate of Parallels. Through a given point only 
one line can be drawn parallel to a given line. 

From this figure it seems quite evident that only one of the lines 
that can be drawn through P can be parallel to 1. 

While this is no proof for the statement, we are ~—- P - 

probably as convinced that the statement is true 
as we should be if a proof were given. ^ ^ 

53. Transversal. A line which cuts two or more lines is 
called a transversal of those lines. 

For example, in the figure below, the line t is a transversal of the 
lines I and V. 


54. Angles made by a Transversal. In the figure given 
below, it is customary to give special names to certain 
angles, as follows: 

ay by c\ d' are called exterior angles; 
a'y b'y Cy are called interior angles; 
d and 6' are called alternate angleSy 
and similarly for c and a'; 

a and a' are called corresponding angleSy and similarly 
for b and b\ for c and c\ and for d and d\ 

Sometimes a and c' are called alternate exterior angles, and similarly 
for h and d'; but when alternate angles are mentioned we ordinarily 
mean alternate interior angles ; that is, we ordinarily mean d and h' 
or c and a', and this should be understood in every case unless the 
contrary is stated. 

PS 









40 


FUNDAMENTAL THEOREMS 


BOOK I 


Proposition 6. Condition of Parallelism 

55. Theorem. When two lines in the same plane are 
cut hy a transversaly if the alternate angles are equals 
the two lines are parallel. 



Given the two lines /, V in the same plane and cut by the 
transversal t so that the alternate Ax and y are equal. 

Prove that I and V are II. 

The plan is to suppose that the lines meet and then to prove that 
this supposition leads to an impossible result. 

Proof. If I and V are not II they will meet if produced. 
Suppose that they meet at P. 

Then y>x. § 50 

An exterior A of a A> either nonadjacent interior Z.. 

But this is impossible, because it is given that y = x. 

Thus the supposition that I and V are not II leads to an 
impossible result, and hence I and V are II. § 51 

56. Indirect Proof. In the above case we have assumed 
the proposition to be false and have shown that this leads 
to an impossible result. We then conclude that the proposi¬ 
tion must be true. Such a proof is called an indirect proof. 

Since the proof excludes all possibilities other than the one stated in 
the proposition, it is also called a proof hy exclusion. It was formerly 
known as the Reductio ad absurdum, the '' reduction to an absurdity.” 










§§ 55-60 PARALLEL LINES 41 

57. Corollary. Two lines in the same plane perpendicular 
to the same line are parallel. 

Draw the figure. What A in the figure are equal and why ? Then 
by what authority can it be said that the lines are II ? 

58. Corollary. Two lines in the same plane parallel to a 
third line are parallel to each other. 

It is given that the lines x and y are both ^_ 

II to the line 1. y _ 

Then if x and y are not II, suppose that 
they meet at P. If this were possible^ how many lines should we have 
through P II to 17 How does § 52 apply ? 

59. Corollary. When two lines in the same plane are cut 
by a transversal, if two corresponding angles are equal or 
if two interior angles on the same side of the transversal 
are supplementary, the lines are parallel. 

Draw the figure and show that if two corresponding A are equal 
or if two interior A on the same side of the transversal are supple¬ 
mentary, two alternate A must be equal, and that § 55 then applies. 

60. Application. In order to draw a line parallel to a 
given line I and passing through a given point P, a drafts¬ 
man often uses a celluloid triangle, as here shown. He 
lays the hypotenuse along the given 
line I, places a ruler r along one of 
the sides, and slides the triangle 
along the ruler until the hypotenuse 
passes through P. He then draws 
a line I' along the hypotenuse. 

Using this construction, draw a line through a given 
point and parallel to a given line. State the authority 
upon which this construction depends. Could another side 
be used instead of the hypotenuse ? Has the side any 
advantage over the hypotenuse? What other instrument 
besides a triangle could be used for this purpose ? 







42 


FUNDAMENTAL THEOREMS 


BOOK I 


Proposition 7. Parallels cut by a Transversal 

61. Theorem. If two parallel lines are cut by a trans¬ 
versal, the alternate angles are equal. 



Given AB and CD, two II lines cut by the transversal XY in 
the points M and N respectively. 

Prove that ZAMN=Z.DNM. 

The plan is to use an indirect proof. 

Proof. Suppose that ZAMN is not equal to ZDNMy but 
that a line PQ through M makes ZPMN = ZDNM. 

Then PQ is II to CD. § 55 

But this is impossible, § 52 

because AB is given as W to CD. 

Hence ZAMN = ZDNM. 

62. Corollary. If two parallel lines are cut by a trans¬ 
versal, the corresponding angles are equal. 

Show that this depends only upon §§ 35 and 61. 

63. Corollary. If two parallel lines are cut by a trans¬ 
versal, the two interior angles on the same side of the 
transversal are supplementary. 

Show that this depends only on § 61 and certain definitions. 

As a special case, if a line is perpendicular to one of two 
parallel lines, it is perpendicular to the other also. 











§§ 61-63 


PARALLEL LINES 


43 


Exercises. Parallel Lines 


1. If two parallel lines are cut by a transversal, the 
alternate exterior angles are equal. 

Exercises which are printed in italics are given as corollaries in 
some textbooks, and should, therefore, be solved by all students. They 
are not, however, essential to the logical sequence of the propositions, 
as they are not used in the proof of subsequent theorems. 

2. This figure shows two parallel lines 
cut by a transversal. Find the values of 
X, y, z, and w, given that a = 73°; given 
that a = 78°. 




3. Cross arms for electric wires are usually at right angles 
to the poles. What properties of parallels are illustrated 
by several cross arms on one pole ? 

4. In this figure three parallel lines 
are cut by two transversals, and certain 
angles are formed as shown. Find the 
values of w, y, z, and x. 

5. A man who is walking southward changes his direc¬ 
tion to northwest. Through how many degrees does he 
turn? If he wishes to walk southward again, through 
how many degrees must he turn ? Draw a figure showing 
the man's course, and state the proposition upon which 
your second answer depends. 


6. In this figure each angle of I\ABC is 60°, 
and two lines have been drawn parallel to the 
base. What can you discover as to the number 
of degrees in each of the other angles ? 

7. Two parallel lines are cut by a transversal so as to 
make the number of degrees in one interior angle 2x and 
the number of degrees in the other interior angle on the 
same side of the transversal a; - 30. Find the value of a;. 











44 


FUNDAMENTAL THEOREMS 


BOOK I 


Proposition 8. Angles with Parallel Arms 

64. Theorem. If two angles have their arms respec¬ 
tively parallel^ and if both pairs of parallels extend 
either in the same direction or in opposite directions 
from the vertices, the angles are equal. 



Given A x and z with arms respectively II and extending 
in the same direction from the vertices, and A a and d with 
arms respectively II and extending in opposite directions. 

Prove that x = z and that a = d. 

The plan is to show that the A in each pair are equal to the same 
Z or to equal A. 

Proof. Produce the arms of x and z, thus forming Ay, 


Then x = y = z, § 62 

Produce the arms of a and d, and suppose that ^ is a 
transversal II to QR and Q 'R thus forming A b and c. 

Then a = 6, § 61 

and h = c = d, § 62 

,\ a = d. Ax. 5 

It should be pointed out to the class that the arms of two angles 


extend in the same direction if the arms are on the same side of a line 
joining the vertices; otherwise they extend in opposite directions. 












§64 


ANGLES WITH PARALLEL ARMS 
Exercises. Review 


45 


1. If two angles have their arms respectively parallel, 
and if one pair of parallels extend in 
the same direction from the vertices 
and the other pair extend in opposite 
directions from the vertices, the angles are supplementary. 

2. A bricklayer often uses the instrument here 
shown for determining whether a wall is vertical. 

When the plumb line lies along a line that is par¬ 
allel to the edge AB, he knows that the wall is 
vertical. State the geometric principle involved. 

3. In Ex. 2 state the principle involved in the 
assertion that the plumb line is perpendicular to 


B 


each line formed by producing the horizontal lines of the 
brickwork. 


4. In order to draw a line perpendicular to a given line 
I and passing through a given point P, a draftsman lays 
one side of his triangle along 
the given line U places a ruler r 
along the hypotenuse, and slides 
the triangle along the ruler until ^ 
the other side passes through P. 

He then draws a line I' along this side. Using this construc¬ 
tion, draw a line through a given point perpendicular to a 
given line. Explain in full. 

5. In this figure, given that M bisects AB and CD, prove 
that AC is II to DB. 

AC is II to DB if what two alternate angles 
are equal? 

These two angles are equal if what two tri¬ 
angles are congruent ? 

These triangles are congruent according to what proposition? 
































46 


FUNDAMENTAL THEOREMS 


BOOK I 


Proposition 9. Sum of the Angles of a Triangle 

65. Theorem. The sum of the three angles of a triangle 
is a straight angle. 



Prove that xzy = a st. 

Lettering the figure as above, the plan is to show that x — x\y = y\ 
and a;' + 2 : + 2 /' = a st. Z. Then it will follow that x-\- z-\-y — Z.. 

Proof. Suppose p to be a line through C II to AB and 
making zia;' and y' as shown. § 52 

Then a;' + 2 : 4 - 2 /'= a st. Z. §12 

But x' = x 

and y' — y- §61 

Substituting x and y for their equals, x' and y\ we have 
x-\-z-\-y = 2 i^i.Z.. Ax. 5 

This proposition is one of the most important in geometry. 

In the first statement in the proof it is evident that Ax. 10 is also 
involved, but such minor statements are usually omitted in proofs. 
The teacher should call attention to them if necessary. 

For students who have never seen this proposition before, it is an 
interesting exercise to infer its truth by cutting off and fitting together 
the three angles of a paper triangle. 










§§65-68 ANGLES OF A TRIANGLE 47 


66. Corollary. An exterior angle of a triangle is equal to 
the sum of the two nonadjacent interior angles. 


For a? + 2 / = ast. Z, 

and ZA+ZC + 2 / = a st. Z. 

ZA +ZC+2/ = ^ + 2/. 
Subtracting y, we have 

ZA+ZC=a;. 



By subtracting Z C we see that ZA = a; — ZC. 


67. Corollary. If two angles and a side of one triangle are 
equal respectively to two angles and the corresponding side 
of another triangle, the triangles are congruent. 

If the A of one are x, y, z, and the A of the other are x, y, z', then 
x-\-y-\-z —A 

and + + = a st. Z, §65 

and hence x-\-y-\-z = x-\-y-^z\ Ax. 5 

z — z'. Ax. 2 


Hence, whatever side is taken, the A are congruent. § 44 


68. Corollary. If the hypotenuse and an adjacent angle 
of one right triangle are equal respectively to the hypote¬ 
nuse and an adjacent angle of another, the triangles are 
congruent. 


Consider the figures here shown, in which A ARC and A'B'C' are 
rt. A with Z A =Z A'and AC=A'C'. 

Since the rt. A are also equal (Post. 6), 
the third A must be equal (§ 65). 

We then have 

AC=A'C', 

ZA = ZA', .. 

and ZC = ZC'. 



Hence the A are congruent by § 44. 

It should be observed that this is really a fourth congruence theorem, 
but it follows so easily from § 65 as to be properly a corollary of this 
proposition. 







48 


FUNDAMENTAL THEOREMS 


BOOK I 


Exercises. Angles of a Triangle 

1. If two triangles have the sum of two angles of one 
equal to the sum of two angles of the other, even though 
the angles themselves are not respectively equal, the third 
angles are equal. 

2. An equiangular triangle is also equilateral. 

3. The sum of the two acute angles of a right triangle 
is 90°. 

4. In a draftsman’s triangle, is a 
right angle, as shown in the figure, and 
ZA is often 30°. In such a triangle how ^ 
many degrees are there in ZC? 

5. If one angle of a right triangle is 37°, what is the 
size of the other acute angle ? 

6. Prove § 65 by using the figure in § 66 and supposing 

that a line is drawn from B II to AC. \ 

7. In this figure, what single angle 

is equal to a + c ? To the sum of what / 
angles is q equal ? To the sum of what 
angles is r equal? From these three ^ ^ 

relations of angles find the number of degrees in p + g + r. 

8. In finding the distance of the moon from the earth 
it is necessary to find first the AAMB 
at the center of the moon, AB being 
the diameter of the earth. Obser¬ 
vations are taken on opposite sides 
of the earth at A and B. The lines 
I, V are II, and A a and h are accurately measured. Show 
how, from a and 6, to find AM. 

Such figures are necessarily distorted. The details of the finding of 
A a and h need not be considered. We simply assume that these 
angles can be measured. 








§§69,70 


ISOSCELES TRIANGLE 


49 


Proposition 10. Equal Sides of a Triangle 

69. Theorem. If a triangle has two equal angles, the 
sides opposite these angles are equal. 



Given the A ABC with Z.B = Z.C, 

Prove that b = c. 

The plan is to prove two A congruent. 

Proof. Suppose that AP is -L to BC. Post. 10 

The sum of the A of AABP is equal to the sum of the 
A of AACP. § 65, Ax. 5 

Then, since AB=AC Given 

and AAPB=AAPC, Post. 6 

because AP is taken as J_ to BC, 

we have ABAP=ACAP. Ax. 2 

AABP is congruent to AACP, § 44 

and hence h = c. § 38 


70. Converse Theorems. It should be observed that § 69 is 
closely related to § 42. When two theorems are so related 
that what is given in one is what is to be proved in the 
other, either theorem is said to be the converse of the other. 

Because a theorem is true it does not always follow that 
its converse is true. 








50 


FUNDAMENTAL THEOREMS 


BOOK I 


Proposition 11. Congruence of Right Triangles 


71. Theorem. If the hypotenuse and a side of one right 
triangle are equal respectively to the hypotenuse and a 
side of anotherj the triangles are congruent. 



Given the rt. k^ABC and A'B'C with hypotenuse AC = 
hypotenuse A^C and with BC = B'C'. 

Prove that A ABC is congruent to AA'B'C\ 

The plan is first to prove that ZA — AA' and then to apply § 68. 

Proof. Place AABC next to AA'B'C' so that BC lies 
along B'C\ B lies on B\ and A and A' lie on opposite sides 


of B'C\ 
Then 

C lies on C\ 

because BC is given equal to B'C\ 

Post. 5 

Also, 

X y = Si st. Z.y 

§12 

and hence 

BA lies along A'R' produced. 

§18 

Since 

AAA'C is isosceles, 

because AC is given equal to A'C', 

§ 19 

we have 

' ZA = ZA'. 

§42 


*. AAB'C' is congruent to AA'B'C\ 

§68 

and 

AABC is congruent to AA'R'C'. 

Ax. 5 

Since the corresponding parts of congruent triangles 
Ax. 5 may be applied to congruence. 

are equal. 













§71 


CONGRUENCE OF RIGHT TRIANGLES 


51 


Exercises. Review 



1. The accuracy of the right angle of a draftsman's 
triangle may be tested by first draw¬ 
ing a line along the side BC with the 
triangle in the position ABC on a 
line A A', and then drawing a line 
along BC with the triangle in the position A!bC, State the 
geometric principle involved. 

2. Given that the arms of these angles 
are respectively parallel, prove that d 





M 


is supplementary to a, to 6, and to c. 

3. The ancient kind of leveling instrument here shown 

consists of an isosceles right triangle. ^ 

When the plumb line cuts the mid¬ 
point M of the base BCy the line BC 
is level. State the geometric principle 
involved. 

4. If a ray of light LP strikes a mirror OP at P, it is 

reflected along a line PP' in such a way that /LQPL = 
Z QPP\ QP being _L to OP. If P' is a point « 

on a mirror OP' which is perpendicular to l 9r— 

the first mirror, the ray is similarly re- lr 

fleeted in a line P'L', QP' being J_ to OP'. P ^ 
Find all the acute angles in the figure in terms of i and 
show that P'L' is II to PL, 

5. Consider Ex. 4 when ZO = 60°; when ZO = 30°. 

6. Prove that if the Js PM, PN from the 
point Pto the sides of an Z-AOB are equal, 
the point P lies on the bisector of Z.AOB, 

Write the general statement of this theorem 
without using letters as is done here. 


^ M " 












52 


FUNDAMENTAL THEOREMS 


BOOK I 


7. A method of finding the distance of a ship off shore 
requires the use of a large wooden isosceles triangle. First 
stand at T and sight along the sides of 
the vertical angle of the triangle to the 
ship S and along the shore on a line TA, 

Then from a point P on TA sight to T 
and S along the sides of a base angle of 
the triangle. Then TP= TS. Explain why 
this is true and show how the distance BS from the shore 
to the ship can be found. 



8. ABCD is a square and M is the midpoint of AB. 

With M as center an arc is drawn, cutting BC at P and 
AD at Q. Prove that AMRP is congruent to ^ 

AMAQ, and write the general statement of q 
this theorem without making use of letters 
as is done here. 

This statement should read, "If an arc drawn ^ ' M ' ^ 

with the midpoint of one side of a square as center 
cuts the two adjacent sides, then the triangles cut off by," and so on. 

9. Prove that if the perpendiculars from the midpoint 

M of the base AB to the sides of the A ABC are equal, 
then Z.A^/.B, What then follows as to q 

the sides AC and PC? Write the gen- 
eral statement of this theorem without 
referring to a special figure. A \ ^ i B 



10. Supposethat OFisAtoOX WithO 
as center an arc is drawn cutting OX at A 
andOFatP. Then with A as center an arc 
is drawn cutting OF at P, and with P as 
center and the same radius an arc is drawn 
cutting OX at Q. Prove that OP=OQ, 
What triangles are congruent by § 71 ? 


F 









§§ 72-75 QUADRILATERALS 53 

72. Quadrilateral. A rectilinear figure of four sides is 
called a quadrilateral. A quadrilateral is called 
a trapezoid if it has two sides parallel; 
a parallelogram if it has the opposite sides parallel. 

If nonparallel sides of a trapezoid are equal, the figure is said to be 
isosceles. In a trapezoid or a parallelogram the side parallel to the base 
is called the upper base, the base being then called the lower base. 



Trapezoid Parallelogram Rectangle Rhombus 


A parallelogram is called 

a rectangle if its angles are all right angles; 
a rhombus if its sides are all equal. 

73. Distance. The length of the line segment from one 
point to another is called the distance between the points. 

The length of the perpendicular from an external point 
to a line is called the distance from the point to the line. 

The length of a perpendicular from one parallel line to 
another is called the distance between the parallels. 

74. Height or Altitude. The length of the perpendicular 
between the bases of a parallelogram or a trapezoid is 
called the height or the altitude of the figure. 

The length of the perpendicular from the vertex of a 
triangle to the base is called the height or the altitude of 
the triangle. 

For brevity the perpendicular itsqlf, instead of its length, is often 
called the altitude. The term " altitude " is commonly used in school; 
the term ''height” is commonly used in ordinary conversation. 

75. Diagonal. The line segment joining two nonconsecu- 
tive vertices of any figure is called a diagonal of the figure. 












54 


FUNDAMENTAL THEOREMS 


BOOK I 


Proposition 12. Opposite Parts of a Parallelogram 

76. Theorem. The opposite sides of a parallelogram 
are equal and the opposite angles are also equal. 



Given the EJABCD. 

Prove that BC = AD and AB = DC, 
and also that AB=AD and /.A =ZC. 

The plan is to prove two A congruent. 

Proof. Draw the diagonal AC. Post. 1 

Since x = x', y — y\ and AC — AC, § 61, Iden. 

then AA5C is congruent to ACDA. §44 

.-. BC=-AD, AB = DC, and ZB = ZD. § 38 

Adding equal A, AA=AC. Ax. 1 


77. Corollary. A diagonal divides a parallelogram into 
two congruent triangles. 

78. Corollary. Segments of parallel lines cut off by par¬ 
allel lines are equal. 

79. Corollary. Two parallel lines are everywhere equally 
distant from each other. 

If AB and CD are II, what can be said of Js 
drawn from any points in AB to CD (§ 78), and 
hence from all points? 


C 


D 














§§76-80 


PARALLELOGRAMS 


55 


Proposition 13. First Criterion for a Parallelogram 

80. Theorem. If the opposite sides of a quadrilateral 
are equal, the figure is a parallelogram. 



Given the quadrilateral ABCD with BC = AD and AB = DC. 

Prove that the quadrilateral ABCD is a O. 

The plan is to prove that x = x' and y — y'hy congruent A, and then 
to apply § 55. 


Proof. Draw 

the diagonal AC. 

Post. 1 

In the two A it must now be shown that x = x' and y = y'. 


Since 

BC = AD 


and 

AB = DC, 

Given 

and since 

AC = AC, 

Iden. 

we see that I^ABC is congruent to /\CDA. 

§47 


x = x'; 

§38 

whence 

AB is II to DC. 

§55 

Also, 

y = y'', 

§38 

whence 

BC is II to AD. 

§55 

Hence the quadrilateral ABCD is a O. 

§72 


The proposition is sometimes stated with reference to convex 
quadrilaterals; but, as stated in § 7, in this book we consider only those 
rectilinear figures in which each of the angles within the figure is less 
than two right angles. 

PS 









56 FUNDAMENTAL THEOREMS book i 

Proposition 14. Second Criterion for a Parallelogram 

81. Theorem. If two sides of a quadrilateral are equal 
and parallel^ the figure is a parallelogram. 



Given the quadrilateral ABCD with AB equal and II to DC. 

Prove that the quadrilateral ABCD is a O. 


The plan is 

to prove that x = x' and y = y', and then to apply § 55. 

Proof. Draw the diagonal AC. 

Post. 1 

Since 

AC = AC, 

Iden. 

since 

AB = DC, 

Given 

and since 

X = x\ 

§61 

we see that 

A ARC is congruent to ACDA. 

§40 

Then 

y' = y- 

§38 


.*. BC is II to AD. 

§55 

Also, 

AB is II to DC. 

Given 


.*. ABCD is a O. 

§72 


82. Corollary. If both pairs of opposite angles of a quadri¬ 
lateral are equal, the figure is a parallelogram. 

The sum of the A of the above quadrilateral is the same as the sum 
of the A of the ^ABC and CDA\ that is, it is 4rt.^ (§66). Now if 
ZA = ZC and AB = AD, it follows (Ax. 1) that AA-\- ZB = AC + ZD; 
whence ZA-\-ZB = ^of4rt.A = 2rt.A. Similarly, Z AZD —2rt A. 

Hence, by § 59, the opposite sides are II, and ABCD is a O (§ 72). 









§§81-84 


PARALLELOGRAMS 


57 


Proposition 15. Diagonals of a Parallelogram 

83. Theorem. The diagonals of a parallelogram bisect 
each other. 



Given the O ABCD with the diagonals AC and BD inter¬ 
secting at O. 

Prove that AO = OC 

and that BO = OD, 

The plan is to show first that AAJ50 is congruent to ACDO or that 
ABCO is congruent to A DAO. 

Proof. In &.ABO and CDO we have 

AB=CD. §76 

The opposite sides of a EJ are equal' • •. 

We also have x = x' and y = y'. § 61 

If two II lines are cut by a transversal, the alternate A are equal. 

A ABO is congruent to ACDO, § 44 

If two A and the included side of one A are equal respectively to 
two A and the included side of another, the A are congruent. 

Hence AO = OC and RO = OD. §38 

84. Corollary. If the diagonals of a quadrilateral bisect 
each other, the quadrilateral is a parallelogram. 

For then A ABO is congruent to ACDO (§40), x = x' (§ 38), and AB 
is II to DC (§ 55). Similarly, AD is II to BC. Give the proof in full. 









58 


FUNDAMENTAL THEOREMS 


BOOK I 


Proposition 16. Parallels intercept Equal Segments 

85. Theorem. If three or more parallels intercept equal 
segments on one transversal, they intercept equal seg¬ 
ments on every transversal. 



Given several II s intercepting the equal segments x, y, z on 
the transversal t and intercepting the segments AB, BC, CD 
on the transversal t\ 

Prove that AB = BC — CD. 

The plan is to prove three A congruent. 

Proof. If t is II to t\ the proposition is true by § 78. 

If t is not II to t\ we can evidently prove the theorem if we can 
show that A BC, CD are sides of congruent A. This can be done by 
§ 44 if we can prove that AP = BQ = CR and can prove that the A 
including these lines are respectively equal in each case. 


Suppose that AP, BQ, CR are each II to t. § 52 

Since AP = x, BQ = y, and CR = z, § 78 

we have AP = BQ = CR. Ax. 5 

Then ZBAP=ZCBQ=ZDCR, §62 

and ZAPB = ZBQC = ZCRD. §64 

Hence AABP, BCQ, CDR are congruent, § 44 

and AB-=BC-=-CD. §38 











§§85-88 PARALLELS INTERCEPTING SEGMENTS 59 


86. Corollary. If a line parallel to one 
bisects another side, it bisects the third 
side also. 

Given the A ABC as shown, with DE II to BC and 
BD = DA. 

Prove that CE = EA. 


side of a triangle 



—-r 



In the proof suppose that XY is II to DE. Then show that this is 
simply a special case of § 85, the two transversals being AB and AC. 

The student will find it interesting to take other special cases,—for 
example, the case in which the transversals cross between the lU. 


87. Corollary. The line which joins the 
sides of a triangle is parallel to the third 
to half the third side. 

Given the A ABC as shown, with BD =DA and 
CE = EA. 

Prove that DE is W to BC 
and that DE =\BC. 

In the proof suppose thatE'Eis II to AB. The corollary is evidently 
proved if we can prove that BE ED is a O and that BE = EC. 

Show that a line from D\\ to BC makes CE = EA. Then what follows 
as to DE and RC? How does EE divide RC? 


midpoints of two 
side and is equal 


A 



88. Corollary. a line parallel to the base of a trapezoid 
bisects one of the other sides, it bisects the opposite side and 
is equal to half the sum of the bases. 

Given the trapezoid ABCD as shown, •' 

PG II to AB and AP = PD. 

Prove that RQ = QC 
and that PQ=\{ABaDC). 

• Proof. Suppose that CX is II to DA. 

Then XY = YC, and RQ = QC. § 85 

Hence FQ=iXR. §87 

Also, PY = AX = DC=\{2AX:)=\{AXaDC). §78 

.-. PY+YQ=l {AX + XB + DC)=l {AB + DC). Axs. 1,5 










60 


FUNDAMENTAL THEOREMS 


BOOK I 


Exercises. Review 


1. In this figure, B, C, and D are in a straight line. If 
X =73°, y = 49°, and z = 58°, prove that 
CE is II to BA and find the number of 
degrees in ZB and in AECA. 

2. In the figure of Ex. 1 suppose that b 
X = 138°, y = 15°, and z = 27°. Prove that 
CE is II to BA and find the number of degrees in ZB. 



The student should sketch a new figure, in which the angles conform 
approximately to the new measurements. ^ 

3. In this figure, ifx = 34°, y = 49°, and 

z = 83°, then AB is II to CB. D 

Produce RQ to meet AB. A 

4. In the figure of Ex. 3, if it is given 

that AB is II to CD, then z — x-\-y, C^ 


5. In this figure. O'A' is ± to OA, and O'B' 
is Z to OB. Name all the pairs of equal 
angles in the figure and prove each statement. 



6. In the figure of Ex. 5, what other condi¬ 
tion would make the two triangles congruent ? 


o 


■A 


7. In Ex. 5 suppose that O' lies within ^AOB, as shown 
in this figure. 

Produce B'O' to meet OA, as at X. Show that 
the angles of AXOA' are respectively equal to 
the angles of AX OB'. 

8. In Ex. 7 prove that Z B'O'A'is supple- 
mentary to Z O. 

9. In Ex. 5 suppose that O' lies on OB, 
as shown in this figure. 

Show that the angles of A B'O A'are respectively 
equal to the angles of A B'O O'. 









§§8»-93 POLYGONS 61 

89. Polygon. A rectilinear figure of three or more sides 
is called a polygon. 

The terms sides, perimeter, angles, vertices, and diagonals are 
employed in the usual sense in connection with polygons in general. 

90^ Polygons classified as to Sides. A polygon is called 

a triangle if it has three sides; 
a quadrilateral if it has four sides; 
a pentagon if it has five sides; 
a hexagon if it has six sides. 

These names are sufficient for most cases. The next few names in 
order are heptagon, octagon, nonagon, decagon, undecagon, dodecagon. 

A polygon is equilateral if all its sides are equal. 

91. Polygons classified as to Angles. A polygon is 

equiangular if all its angles are equal; 

convex if each of its angles is less than a straight angle; 

concave if it has an angle greater than a straight angle. 



Equilateral Equiangular Hexagon Convex Concave 


In a concave polygon, an angle greater than a straight angle is 
called a reentrant angle. As stated in § 7, when the term polygon 
is used a convex polygon is understood unless the contrary is stated. 

92. Regular Polygon. A polygon that is both equiangular 
and equilateral is called a regular polygon, 

93. Relation of Two Polygons. Two polygons are 
mutually equiangular if the angles of the one are equal 

to the angles of the other, taken in the same order; 

mutually equilateral if the sides of the one are equal to 
the sides of the other, taken in the same order. 





62 


FUNDAMENTAL THEOREMS 


BOOK I 


Proposition 17. Sum of the Angles, of a Polygon 

94. Theorem. The sum of the interior angles of a 
^polygon is as many straight angles less two as the 
figure has sides. 



Given the polygon ABCDEF with n sides. 

Prove that the sum of the interior A is {n — 2) st. A. 

The plan is to cut the figure into A and apply § 65. 

Proof. From any vertex A draw as many diagonals as 
possible. Then there is a A for each side except the two 
adjacent to A. Hence there are (n — 2) A. 

The sum of the A of each A is a st. Z. § 65 

Hence the sum of the A of the (72 —2) A, that is, the 
sum of the A of the polygon, is (n — 2) st. A. Ax. 3 

Notice that this proposition includes § 65 as a special case. 

95. Corollary. The sum of the angles of a quadrilateral is 
two straight angles; and if the angles are all equal, each is 
a right angle. 

Give brief oral proofs of all such corollaries. 

96. Corollary. Each angle of a regular polygon of n sides 
is equal to {n —2)/n straight angles. 










§§94-96 


ANGLES OF A POLYGON 


63 


Exercises. Review 

1. If the arms of one angle are respectively perpendicular 
to the arms of another angle, the angles are either equal or 
supplementary. 

2. Any two consecutive angles of a parallelogram are 
supplementary. 

3. If one angle of a triangle is 37° 30', what is the sum 
of the other two angles ? 

4. If the sum of two angles of a triangle is 37° 30', how 
many degrees are there in the other angle ? 

5. If an exterior angle at the base of an isosceles tri¬ 
angle is 98°, find the number of degrees in each angle of 
the triangle. 

6. If the exterior angle at the vertex of an isosceles 
triangle is 98°, find the number of degrees in each angle 
of the triangle. 

7. In this figure, which shows two parallel 
lines cut by a transversal, x = 59°. How many 
degrees in each of the other seven angles? 

8. Find the sum of the angles at the five points of the 
usual form of the five-pointed star. 

Such a star is sometimes called a 'pentagram. It 
was used as a badge by the followers of Pythagoras, q., 
one of the greatest of the Greek mathematicians, 
about 525 b.c. At the five points were the Greek 
letters v, 7 , t, e, a, the word vyieia (hygieia) mean¬ 
ing ’’health,’' the single letter e being used for ei. 

9. Study this figure with respect to the 
sum of the marked angles, write a theorem 
concerning it, and prove this theorem. 

10. Consider the theorem of Ex. 9 for 
the special case of the parallelogram. 









64 FUNDAMENTAL THEOREMS book i 

Proposition 18. Exterior Angles 

97. Theorem. The sum of the exterior angles of a 
polygon, made by producing each of its sides in suc¬ 
cession, is two straight angles. 



Given the polygon ABCDEF with its n sides produced in 
succession. 

Prove that the sum of the exterior A is 2 st. A. 

The plan is to take the sum of the interior A from n st. A. 

Proof. Designate the interior A by a, h, c, d, e,f, and the 
corresponding exterior A by a', b', c', d\ e\ f. 

Then, considering each pair of adjacent A, 
a-\-a'= a st. Z, 

and 6 + 6'=ast. Z. §12 

In like manner, each pair of adjacent Z = a st. Z. 

Then, since the polygon has n sides and n A, the sum of 
the interior and exterior Z is st. Z. Ax. 3 

But the sum of the interior Z is (n — 2) st. Z § 94 

or st. Z — 2 st. Z. 

Hence st. Z — (n st. Z — 2 st. Z) = 2 st. Z; 
that is, the sum of the exterior Z is 2 st. Z. 


Ax. 2 








§97 


EXTERIOR ANGLES 


65 


Exercises. Review 

1. In making a map of a field a surveyor uses an in¬ 
strument which enables him to find with equal ease the 
interior angles and the exterior angles 
of the field. In order to check his work 
he may use either § 94 or § 97. Which is 
the easier for him to use, and why is it 
easier? 

2. In making a map of a field of five sides a surveyor 
finds that the exterior angles are 20° 30', 39° 30', 59° 30', 
35° 30', and 24° 30'. Are his angle measures correct ? If all 
but the last are checked and thus are known to be correct, 
what is the size of the last angle ? 

3. This figure represents two pairs of parallel lines. 
State all the equalities of angles, thus: 
a = c = g = e = o = *r». Give the reason 
in each case. 

4. In the figure of Ex. 3 state ten 
pairs of nonadjacent angles which are 
supplementary; thus: a-f/i = 180° and 
d + e=lS0\ 

5. In this figure, given that AC = BC 
and that DE is II to AB, prove that 
CD = CE. Write a general statement of 
the theorem. 

6. In the figure here shown, a; = 72° 
and x = ly. Is II to CD ? Give the 
proof in full. 

7. In the figure of Ex. 6 suppose that 
X = 73° and y-x = 32°. Is AB then II to CD ? Give the proof. 

8. How many sides has a regular polygon each angle of 
which is 140° ? 











66 


FUNDAMENTAL THEOREMS 


BOOK I 


98. Summary of Important Fundamental Theorems. There 
are many important theorems in geometry, but those 
which we have thus far studied are used more often than 
those of any other similar group. We may now summarize 
the most important of the results as follows: 

Conditions of Congruence of Triangles 

1. Two sides and included Z respectively equal. § 40 

2 . Two A and included side respectively equal. § 44 

3. Three sides respectively equal. § 47 

4. Two A and any side respectively equal. § 67 

Conditions of Congruence of Right Triangles 

1. Hypotenuse and an adjacent Z respectively equal. § 68 

2. Hypotenuse and a side respectively equal. § 71 


Conditions of Parallelism 


1. Alternate A equal. 

2. Two lines J_ to the same line. 

3. Two lines II to a third line. 

4. Corresponding A equal. 

5. Interior A on same side supplementary. 


§55 

§57 

§58 

§59 

§59 


Transversal Cutting Parallels 


1. Alternate A are equal. 

2. Corresponding A are equal. 

3. Interior A on same side are supplementary. 

4. Segments on other transversals are equal. 


§61 

§62 

§63 

§85 


Sums of Angles 


1. Of a triangle, 

2. Of a polygon, 

3. Of a polygon, exterior. 


{n — 2) st. Z. 
2 st.Z. 


lst.Z. 


§65 

§94 

§97 


FUNDAMENTAL CONSTRUCTIONS 


67 


II. Fundamental Constructions 

99. Construction. When we construct a figure we make 
the figure accurately by the aid of an unmarked ruler and 
a pair of compasses, which are the only instruments recog¬ 
nized in elementary geometry. When we draw a figure we 
make the figure without the aid of these instruments, but- 
we may use, if we wish, the draftsman’s triangle, the pro¬ 
tractor, or the T-square, so as to make a neat figure. 

In many cases it is immaterial whether we use the word draw ” 
or the word construct,” as when we speak of drawing a line. 

We shall now consider the solution of a few of the most important 
problems of construction. 

100. Nature of a Solution. A solution of a problem has 
one step that a proof of a theorem does not have. 

In proving a theorem we state (1) what is given, (2) what 
is to be proved, and (3) the proof. 

In solving a problem we must state (1) what is given; 

(2) what is required, that is, to do some definite thing; 

(3) the construction, that is, how to do it; and (4) the proof, 
showing that the construction explained in step 3 is correct. 

We prove a theorem, but we solve a problem and then 
prove that our solution is a correct one. 

In the figures for the problems in Book I, given lines are shown as 
full lines, required lines as heavy black lines, and construction lines and 
lines produced as dotted lines. (See also the note in § 39.) 

101. Discussion of a Problem. Besides the four necessary 
steps mentioned in § 100, a fifth step may profitably be 
taken in connection with every problem. This step is the 
discussion of the solution, to see if there are any interesting 
special cases in which a solution is impossible or in which 
there is more than one solution. Such discussions are, in 
general, left to the teacher and students. 


68 


FUNDAMENTAL CONSTRUCTIONS 


BOOK I 


Proposition 19. Bisecting a Line Segment 
102. Problem. Bisect a given line segment. 


X 

.L 

"/ i W 

1 \ \ 
m\ I '\p 

/ 

/ 

1/ 

i i Z 

i / 

\ 1 

Y 


Given the line segment AB, 

Required to bisect AB. 

The plan is to construct two congruent A. 

Construction. With A and B as centers and with any con¬ 
venient radius construct two arcs that intersect. Post. 4 

A convenient radius in many cases is AB itself. 

Designate the points of intersection of the arcs asXand Y. 
Draw the st. line XY and designate the point where it 


cuts the given line segment as M. Post. 1 

Then XY bisects AB at M ; that is, AM—BM. 

Proof. Draw AX, BX, AY, BY. Post. 1 

- Since AAYX is congruent to A RYX, §47 

we have Z Z BXY. § 38 

.*. AAMX is congruent to /\BMX. § 40 


The student has here the essential features of the proof. He should 
DOW give the steps in full. 











§§ 102,103 


BISECTIONS 


69 


Proposition 20. Bisecting an Angle 
103. Problem. Bisect a given angle. 



Given the Z.AOB. 

Required to bisect /LAOB. 

The plan is to construct two congruent A. 

Construction. With O as center and any convenient radius 
describe an arc cutting OA at X and OB at Y. Post. 4 

With X and Y as centers and with a radius greater than 
half the line segment from X to F, construct intersecting 
arcs and designate their point of intersection as P. Post. 4 

A convenient radius may be found by placing one point of the 
compasses on X and the other on F. 


Draw 


OP. 

Post. 1 

Then 


OP bisects ZAOB. 


Proof. Draw 

PX and PY. 

Post. 1 

Since 


OX= OY, 

Post. 4 

since 


PX=PY, 

Const. 

and since 


OP=OP, 

Iden. 

we see that 

AOXP is congruent to AOYP. 

§47 



ZXOP = ZYOP. 

§38 








70 


FUNDAMENTAL CONSTRUCTIONS 


BOOK I 


Proposition 21. Perpendicular through Internal Point 

104. Problem. Through a given point on a given 
straight line construct a perpendicular to the line. 



Given the line AB and the point P on AB. 

Required through P to construct a 1. to AB. 

The plan is to construct two congruent A. 

Construction. By drawing arcs, make PX=PY. Post. 4 
With X as center and XY as radius construct an arc, and 
with Y as center and the same radius construct another 
arc intersecting the first arc at C. Post. 4 

Draw PC, which is the required J_. Post. 1 

Proof. Draw CX and CY. Post. 1 

Since we used the same radius in constructing the inter¬ 
secting arcs, we have 

CX= CY. Const. 

Also PX=PY, Post. 4 

and CP = CP. Idem 

AXPC is congruent to AYPC, § 47 

and ZCPX=ZCPY. §38 

.‘.ZCPX is a rtZ, §13 

PC is _L to AP. §14 


and 









§§ 104, 105 


PERPENDICULARS 


71 


Proposition 22. Perpendicular through External Point 

105. Problem. Through a given point outside a given 
straight line construct a perpendicular to the line. 



Given the line AB and the point P not on AB, 

The plan is to construct two congruent A. 

Required through P to construct a A- to AB, 

Construction. With P as center and a radius sufficiently 
long construct an arc cutting AB at X and Y, Post. 4 

Such a radius can easily be found by simply placing one point of the 
compasses on P and the other on any point below AB. 

With X and Y as centers and a radius sufficiently long 


construct two arcs intersecting at C below AB, Post. 4 

Such a radius may be any length greater than half of XY. 

Draw PC, Post. 1 

Let M be the point of intersection of PC and AB, 

Then PM is the required ±. 

Proof. Draw PX, PF, CX, CY, Post. 1 

Then APXC is congruent to APYC, § 47 


Now write out the full proof, which should show that AXMP is 
congruent to A YMP by § 40. 

PS 









72 


FUNDAMENTAL CONSTRUCTIONS 


BOOK I 


Exercises. Constructions 

1. Draw a line segment 3} in. long and bisect this line 
segment by measuring. Then bisect it by § 102 and thus 
test the accuracy of your measurement. 

2. By the aid of a protractor draw and bisect an angle 
of 60°. Then bisect the angle by § 103. 

3. Draw a line AB, take a point P not on AB, and through 
Pdraw a perpendicular to AB by means of a draftsman's 
triangle. Then through P construct a perpendicular to AB 
by the method of § 105, and thus check the accuracy of the 
drawing. 

In ordinary practice, either of these methods is satisfactory. 

4. Draw a line AB, take a point P on the line, and through 
P draw a perpendicular to AB by means of a draftsman's 
triangle. Then construct a perpendicular as in § 104. 

5. Write a statement about the relative sizes of the halves 
of equal line segments; of the halves of equal angles; of 
the halves of equal circles; of the halves of any equal 
magnitudes. Draw a diagram to illustrate each statement. 

6. Write a statement about the result of adding equal 
line segments to equal line segments; of adding equal 
angles to equal angles. Draw a diagram to illustrate each 
statement. 

7. How many degrees are there in an angle that is equal 
to half its complement? to half its supplement? 

8. How many degrees are there in an angle that is equal 
to 10° more than its complement ? to 20° less than its com¬ 
plement? to 30° less than half its complement? 

9. Construct a line segment equal to the sum of two 
given line segments; to the difference between two given 
line segments. 


§105 


EXERCISES 


73 


Construct angles of the following sizes: 

10. 45°. 11. 22°30'. 12. 11° 15'. 13. 135°. 14. 157°30'. 

15. Construct a square 2 in. on a side. If the figure is cor¬ 
rectly constructed the two diagonals are equal. Check the 
work by measuring the diagonals with the compasses. 


16. By the use of 
following figures: 



compasses and ruler construct the 



The lines made of short dashes show how to locate the points needed 
in drawing the figure. They should be erased after the figure is com¬ 
pleted unless the teacher directs that they be retained to show how 
the construction was made. 

17. By the use of compasses and ruler construct the 
following figures: 



In the figures in Exs. 16 and 17 it should be noticed that the radius 
of a circle may be used to draw arcs which shall divide the circle into 
six equal parts. 

18. By the use of compasses and ruler construct four 
original designs similar in nature to those of Ex, 17, Try 
to make the designs as varied as possible. 











74 


FUNDAMENTAL CONSTRUCTIONS 


BOOK I 


Proposition 23. Constructing Equal Angles 

106. Problem. From a given point on a given line con¬ 
struct a line which shall make with the given line an 
angle equal to a given angle. 



Given the AAOB and the point P on the line ^1?. 

Required from P to draw a line making with the line QR 
an Z equal to /LAOB. 

The plan is to construct two congruent A. 

Construction. With O as center and any radius describe 


an arc cutting OA at C and OB at D. Post. 4 

With P as center and the same radius describe an arc 
MX, cutting QR at M. Post. 4 

Draw CD. Post. 1 

With M as center and CD as radius describe an arc cutting 
the arc MX at N. Post. 4 

Draw PN. Post. 1 

Then PN is the required line. 

Proof. Draw MN. Post. 1 


Now prove that A OCD and PMN are congruent by § 47. 

This method of constructing equal angles is more nearly accurate 
than the method of drawing by the aid of a protractor. 









§§ 106-109 CONSTRUCTING EQUAL ANGLES 75 

107. Corollary. Through a given external point construct 

a line parallel to a given line. x 

Let P be the given external point and AB the '' \ , 

given line. C . D 

Draw any line XPY through P, cutting AB as / \\ ’ 

in the figure. ^_ /v ^ 

At P construct p = q, and draw DPC. ^ 

The line CD is the required line. 

Write the construction in the usual form and give the proof. 

108. Corollary. Given two sides and the included angle 
of a triangle, construct the triangle. 

Let h and c be the given sides 
and m the given Z. 

Construct ZXOY= m. 

On OX mark off with the com¬ 
passes OB=c, and on OF mark 
off OC= b. 

Draw BC. 

Then A OBC is the required A. 

Write the construction in the usual form and give the proof. 

Of course the A may be turned over, giving another appearance, 
but such cases, if thought important, are left to the consideration of 
the class. 

109. Corollary. Given a side and two a,ngles of a triangle, 
construct the triangle. 

Let a be the given side and 
m and n the given A. 

Then if the side is included by 
the A, mark off with the com¬ 
passes on any line I the segment 
PQ = a. 

At P construct an Z equal to w, and at Q construct an Z equal to n. 

Then APQR in the figure is the required A. 

Write the construction in the usual form and give the proof. 

If the side is not included by the A, find the third Z by means of 
§ 65 and then proceed as above. 















76 


FUNDAMENTAL CONSTRUCTIONS 


BOOK I 


Proposition 24. Triangle with Given Sides 

110. Problem. Construct a triangle with its sides 
equal respectively to three given line segments. 



Given the line segments /, m, n. 

Required to construct a A with sides equal to Z, m, n. 

The plan is to draw two arcs which shall determine the A. 

Construction. Draw a line a with the ruler and on it 
mark off with the compasses a line segment AB = Z. 

With A as center and m as radius draw an arc; with B 
as center and n as radius draw another arc cutting the first 


arc at C. 


Post. 4 

Draw 

AC and BC. 

Post. 1 

Then 

ABC is the required A. 


Proof. 

AB = Z, AC = m, and BC = n. 

Const. 


The discussion (§ 101) should disclose any special cases. 

111. Corollary. Given one of the sides, construct an equi¬ 
lateral triangle. 

In this case, and similarly in § 112, the student should perform the 
construction, and then write out the construction and the proof in 
proper geometric form. 

112. Corollary. Given the base and one of the two equal 
sides, construct an isosceles triangle. 












§§ 110-113 


DIVIDING A LINE 


77 


Proposition 25. Dividing a Line Segment 

113. Problem. Divide a given line segment into a given 
number of equal parts. 



Given the line segment AB, 

Required to divide ABinto a given number of equal parts. 
The only proposition thus far studied that relates to equal segments 
on a line is the one concerning a transversal cutting lls (§ 86). The plan 
is, therefore, to bring this problem under that theorem. 

Construction. From A draw the line AX, making any con¬ 
venient Z with AB. Post. 1 

Take any convenient length and, by describing arcs, apply 
it to AX as many times as is indicated by the number of 
parts (say three) into which AB is to be divided. Post. 4 
From R, the last point thus found, draw RB. Post. 1 
From the points P, Q by which AX was divided into 
equal parts, construct FF' and QQ' II to RB. § 107 

These lines divide AB into equal parts as required. 

Proof. Construct AY W to BR. §107 

Since the lls AY, F'F, Q'Q, BR were constructed so as to 
cut off equal segments on AX, they cut off the equal seg¬ 
ments AF\ F'Q', Q’B on AB. § 85 

This method is more nearly accurate than trying to divide AB by 
measuring its length with a ruler. 








78 


FUNDAMENTAL CONSTRUCTIONS 


BOOK 


Proposition 26. Two Sides and One Angle 

114. Problem. Given two sides of a triangle and the 
angle opposite one of therriy construct the triangle. 



Given a and two sides of a A, and A the Z opposite a. 

Required to construct the A. 

The plan is to determine the A by means of arcs. 


Construction. CASE 1. If a <b. 

On a line AX construct XXA r=Z A. § 106 


On A F take AC =b. Post. 4 


With C as center and a as radius construct an arc inter¬ 
secting the line AX at B and B\ Post. 4 

Draw BC and P'C, thus completing the A. Post. 1 

Then both A ARC and AAB'C satisfy the conditions. 

This is called the ambiguous case. 


Except for students specializing in mathematics, 
omitted. 

For the present we shall assume 
that if a < 6 there are, in general, 
two constructions as stated. If a 
is equal to the ± from C to AX, 
it is evident that there is but 
one construction, the rt. A ARC, as shown in the figure at 
If a is less than the ± from C to AX, it is apparent that 
no A, as shown in the figure at the right. 



the left, 
there is 















§114 


CONSTRUCTION OF A TRIANGLE 


79 







Case 2. Ifa = b, 

If the given ZA is acute and a = 6, the arc constructed 
from C as center with radius a apparently cuts the line 
WX at the points A and B. There 
is, however, but one A; namely, the 
isosceles AABC. 

If A is a rt. Z or an obtuse Z, there is 
no A when a = b, for a A cannot have two 
rt. A or two obtuse A (§ 65). 

Case 3. If a>b. 

If the given Z A is acute, the arc constructed from C cuts 
the line WX on opposite sides ^ 

of A at the points B and B', 

Then AABC satisfies the con¬ 
ditions, but AAB'C does not, 
for it does not contain the acute 
ZA. There is then only one A 
that satisfies the conditions. 

If the given ZA is a rt. Z, 
the arc constructed from C cuts 
the line WX on opposite sides of 
A at the points B and B\ and 
we have two congruent rt. A 
that satisfy the conditions. 

If the given ZA is obtuse, the arc constructed from C 
cuts the line WX on opposite 
sides of A at the points B and 
B'; but only the AABC satisfies 
the conditions. 

The proofs of these statements are 
given later, but since this proposi¬ 
tion will not be used in proving any 
theorems, it is permissible to use them here in discussing the problem. 









80 


FUNDAMENTAL CONSTRUCTIONS 


BOOK I 


Exercises. Review of Constructions 

1. Divide a given line segment into four equal parts. 

2. Construct an equilateral triangle of given perimeter. 

3. Through a given point draw a line which shall make 
equal angles with the two sides of a given angle. 

4. Through a given point draw two lines which shall 
form with two intersecting lines two isosceles triangles. 

5. Construct a triangle with its three angles respec¬ 
tively equal to the three angles of a given triangle. 

By first constructing an equilateral triangle and then 
bisecting certain angles construct angles of: 

6.. 30°. 7. 15°. 8. 7° 30'. 9. i of a rt. Z. 

10. Construct an isosceles triangle with its base equal 
to one third of one of the equal sides. 

11. Construct an isosceles right triangle. 

12. Construct an isosceles triangle with one of the base 
angles 60°. What other special name can you give to the 
triangle ? Prove that your answer is correct. 

13. By the use of compasses and ruler construct the 
following figures (see Ex. 16, page 73): 



In such figures artistic patterns may be made by coloring various 
portions of the drawings. In this way designs are made for oilcloth, 
for stained-glass windows, for colored tiles, and for other decorations. 



§115 


INEQUALITIES 


81 


III. Inequalities 

Proposition 27. Unequal Sums of Lines 

115. Theorem. The sum of two line segments from a 
given external point to the extremities of a given line 
segment is greater than the sum of two other line seg¬ 
ments similarly drawn hut included hy them. 



Given the line segment AB and the segments from the ex¬ 
ternal points Qy P to A and B, 

Prove that AQ-\- QB>AP-{-PB. 

The plan is to show that AQ-\- QB > AX + XB > AP + PB. 


Proof. Produce AP to meet QB as at X, Post. 2 

Then AQ + QX>AP-fPX Post. 3 

Likewise, PX+XB>PB, Post. 3 

Adding these inequalities, we have 

AQ + QX+PX+XP>AP+PX+PP. Ax. 8 

Substituting QB for its equal, QX + XB, we have 

AQ + QP + PX>AP+PX+PP. Ax. 5 

.*. AQ-f QP>AP+PP. Ax. 7 


It may be asked why AP produced meets BQ at any point whatever. 
Such discussions, of little significance at this stage, are left to the teacher 
to initiate if thought desirable. 








82 


INEQUALITIES 


BOOK I 


Proposition 28. Perpendicular from an External Point 

116. Theorem. One and only one perpendicular can he 
constructed to a given line from a given external point. 



Given a line XY and an external point P. 

Prove that one and only one ± can he constimcted from 
P to XY. 

The plan is to show that if two lines from P are _L to XY, then 


Post. 1 is violated. 

Proof. One 1. to XY, as PO, can be constructed. § 105 
Let PZ be any other line from P to XY. Post. 1 

Produce PO to P', making OP' = OP. Post. 2 

Draw P'Z. Post. 1 

Since POP' is a st. line, PZP' is not a st. line. Post. 1 
Hence ZLP'ZP is not a st. Z. § 12 

Since APOZ and P'OZ are rt. Z, § 14 

we have APOZ = ZP'OZ Post. 6 

Hence AOPZ is congruent to AOP'Z, § 40 

so that Z OZP = Z OZP'. § 38 

.*. AOZP, the half of AP'ZP, is not a rt. Z. § 13 
Hence PZ is not ± to XY, § 14 

and PO is the only Z to XY. 

We may now cease to depend upon part of Post. 10. 








§§116.117 PERPENDICULARS AND OBLIQUES 


83 


Proposition 29. A Perpendicular and Equal Obliques 

117. Theorem. If two line segments drawn from a point 
on a perpendicular to a given line cut off on the given 
line equal segments from the foot of the perpendicular, 
the line segments are equal and make equal angles with 
the perpendicular. 



Given PO J_ to XF, and PA and PB two lines cutting off 
from O on XY the equal segments OA and OB, 

Prove that PA = PB, 

and that AAPO = X.BPO, 

The plan is to prove that the AAOP and BOP are congruent. 


Proof. Since PO is ± to XY, Given 

we see that APOA and FOB are rt. § 14 

APOA = APOB, * Post. 6 

Also, OA = OB, Given 

and PO = PO, Iden. 

Hence AAOP is congruent to A BOP, § 40 

,\PA = PB, 

and AAPO=ABPO, §38 


While not dealing directly with inequalities, §§ 116 and 117 are related 
to the theory, as is shown later. 









84 


INEQUALITIES 


BOOK I 


Proposition 30. A Perpendicular and Unequal Obliques 

118. Theorem. If two line segments drawn from a 
point on a perpendicular to a given line cut off on the 
given line unequal segments from the foot of the perpen¬ 
dicular, the line segment more remote is the greater. 



Given PO _L to XY and two lines PA, PC drawn from P to 
XY so that OA > OC. 


Prove that 

PA>PC. 


The plan is to show that PA > PB, which is equal to PC. 


Proof. Take 

OB =OC and draw PB. 

Post. 1 

Then 

PB = PC. 

§117 

Produce 

PO to P\ making OP'= OP. 

Post. 2 

Draw 

P'A and P'B. 

Post. 1 

Then 

PA = P'A and PB = P'B. 

§117 

But 

PA+P'A>PB + P'B, 

§115 

because PP' 

is a line segment to the ends of which we have 
drawn segments from A and B. 


:.2PA>2PB, 

Ax. 5 

because we may substitute PA for P'A, and PB for P'B. 

Hence 

PA>PB, 

Ax. 7 

and 

PA>PC. 

Ax. 5 









118-121 PERPENDICULARS AND OBLIQUES 


85 


119. Corollary. Only two equal obliques can he drawn froTU 
a given point to a given line. 

Let PA, PB, PC be three obliques and let 
PO be ± to XY. 

Then to suppose that PA = PB = PC is to 
contradict § 118, where it was proved that 
PA>PC. 

120. Corollary. Equal obliques from a point to a line cut 

off equal segments from the foot of the p 

perpendicular from the point to the line. 


JK 

A B O C 


Given PO ± to XY and PA = PB. . 


\ 


Prove that 

OA = OB. / 



\ 

Proof. 

^_ e. _ 

In AAOP and BOP we see that 

C 

) 




APOA and POB are rt. A, 



§14 



because PO is given as A- to XY. 






.•. AAOP and BOP are rt. A. 



§20 

Also, 


PA=PB, 



Given 

and 


PO = PO. 



Iden. 



.*. A A OP is congruent to ABOP, 



§71 

and 


OA = OB. 



§38 


121. Corollary. If two unequal line segments are drawn 
from a point to a line, the greater cuts off the greater seg¬ 
ment from the foot of the perpendic¬ 
ular from the point to the line. 

In this figure, in which PO is _L to XY and 
PA >PB, it is impossible that A should lie 
between B and O. For if A should be at A', 
then PA (that is, PA') would be less than PB 
(§ 118), which is contrary to what is given. Further, A cannot fall on 
B, for then PA = PB, which is also contrary to what is given. 

Thus A cannot lie on B or between B and O. Hence the greater 
segment PA cuts off the greater segment on XY from O. 

Similarly, if PA lies on the right of PO, as at PC, then, since 
PA = PC, we see that OA = OC (§ 120), so that OC> OB. 

Since we have covered all possible cases, the corollary is true. 









86 INEQUALITIES book i 

Proposition 31. Perpendicular Shortest Line 

122. Theorem. The 'perpendicular is the shortest line 
segment that can be constructed to a given line from a 
given external point 



Given PO, the J_ from an external point P to the line XY, 

Prove that PO is the shortest line from P to XY. 

The plan is to show that PO is shorter than any other line. 


Proof. Let PZ be any other line segment from P to XY. 


Produce 

PO to P\ making OP' = OP. 

Post. 2 

Draw 

P'Z. 

Post. 1 

Since XF is given _L to PP\ then PZ = P'Z. 

§117 

Then 

PZ+P’Z=2PZ, 


and 

P0 + P'0 = 2P0. 

Axs. 5,10 

But 

P0 + P'0<PZ+P'Z. 

Post. 3 

Hence 

2PO<2PZ, 

Ax. 5 

and 

PO<PZ; 

Ax. 7 

that is, 

PO is the shortest line from P to XY. 



123. Corollary. Conversely, the shortest line segment to a 
given line from an external point is the perpendicular from 
the point to the line. 

For if PO is the shortest line segment, it must be -L to XY. Other¬ 
wise we should have a line segment from P to XY shorter than the _L, 
which is impossible (§ 122). 








§§ 122-124 


ANGLES OF A TRIANGLE 


87 


Proposition 32. Angles of a Triangle 

124. Theorem. If two sides of a triangle are unequal^ 
the angles opposite these sides are unequaly and the angle 
opposite the greater side is the greater. 



Given the AA5C with CB > CA. 

Prove that /LBAC>Z-B. 

The plan is to show that ZBAC> q = r> ZB. 

Proof. Because CB>CA we may suppose that CX can 
be marked off with the compasses on CB so that CX = CA. 


Draw AX .' Post. 1 

Then AAXC is isosceles. § 19 

Then, in the figure, ^ = r, § 42 

because in an isosceles A the A opposite the equal sides are equal. 

But r>XBy §50 

because an exterior Z of a A'> either nonadjacent interior Z. 

Also, ZBAOq. Ax. 10 

Substituting r for its equal, q, we have 

ZBAC>r. Ax. 5 

Since r>ZB, Proved 

then ZBAOZB. Ax. 9 


If the first of three quantities > the second, and the second > the 
third, then the first > the third. 


PS 








88 


INEQUALITIES 


BOOK I 


Proposition 33. Sides of a Triangle 

125. Theorem. If two angles of a triangle are unequal, 
the sides opposite these angles are unequal, and the side 
opposite the greater angle is the greater. 



Given the l^ABC with Z.A> /LB. 


Prove that a>b. 

The plan is to show that other suppositions lead to an impossibility. 


Proof. Now a is either equal to b, less than b, or greater 
than b. 


If 
then 
And if 
that is, if 
then 


a = b, 
/A=/B. 
a<b, 
b>a, 

/B>/A. 


§42 

§124 


Both these conclusions are contrary to the fact that 

/A>/B. Given 

Hence it follows that a > 6. 


This is another example of an indirect proof (§ 56). We suppose that 
the statement to be proved is false, that is, that a = 6 and that 6 > a, 
and we show that these suppositions lead to impossibilities; namely, 
that ZA=ZB or ZB > ZA, when we know that ZA> ZB. Accord¬ 
ingly, we conclude that the theorem is true. 








§125 


89 


INEQUALITIES IN TRIANGLES 


Exercises. Inequalities 


1. The sum of any two sides of a triangle is greater than 
the third side, and the difference between any two sides is 
less than the third side. 

Use Post. 3 for the first statement and Ax. 7 for the second. 

State in what cases it is possible to form triangles with 
rods of the following lengths, and give the reason: 

2. 2 in., 3 in., 4 in. 5. 7 in., 10 in., 20 in. 

3. 3 in., 4 in., 7 in. 6. 8 in., in., 18 in. 

4. 6 in., 7 in., 9 in. 7. 9| in., lOJ in., 20 in. 

8. In this figure prove thSit AB-^BC> ADDC. 

Why is DB-\-BC> DC": What is the result of 

adding AD to these unequals ? ^ 

9. In the figure of Ex. 8 suppose that 



CA = CB, and prove that CD < CB. Write 
a theorem based upon this fact. 

The theorem may begin as follows: The line segment joining the 
vertex of an isosceles triangle to any point on the base is less than.... 

10. The hypotenuse of a right triangle is greater than 
either of the other sides. 

11. Prove § 122 by the use of § 125. Is this legitimate ? 

It is legitimate in case § 122 was not used directly or indirectly in 
the proof of § 125; otherwise it is not legitimate. 

C 

12. In this figure, given that x is an ob¬ 
tuse angle and that M is the midpoint of 



AB, prove that a < 6. 




A‘ 


M 


Draw a perpendicular from C to AB. 

13. On the base AB of a quadrilateral ABCD the point 
P is taken. Prove that the perimeter of the quadrilateral 
is greater than the perimeter of A PCD. 




90 


INEQUALITIES 


BOOK I 


Proposition 34. Unequal Angles of Triangles 

126. Theorem. If two sides of one triangle are equal 
respectively to two sides of another, hut the included 
angle of the first triangle is greater than the included 
angle of the second, then the third side of the first is 
greater than the third side of the second. 



Given the A ABC and XYZ with b = y, a = x, and ZOZ.Z. 
Prove that c^z. 

In the figure, the plan is to show that AP+PB=AP+PF> z. 

Proof. Place the A so that Z coincides with C, y lies 


along 6, and Y lies on the same side of AC as B. Post. 5 
Then since y = b,X lies on A, and since Z.Z<Z.C, x lies 
within Z.ACB. 

Let CP bisect Z YCB and draw FP. Posts. 8,1 

Then, since CP = CP, CF is given equal to CB, and Z FCB 
is bisected, we see that 

APYC is congruent to APBC. § 40 

.\PY=PB. §38 

Now AP + PF>AF. Post. 3 

.•.AP + PB>AF, Ax. 5 

AB>AY, or c>z. Ax. 10 


and hence 









§§ 126,127 


INEQUALITIES IN TRIANGLES 


91 


Proposition 35. Unequal Sides of Triangles 

127. Theorem. If two sides of one triangle are equal 
respectively to two sides of another^ hut the third side of 
the first triangle is greater than the third side of the 
second, then the angle opposite the third side of the first 
is greater than the angle opposite the third side of 
the second. 



Given the A ABC and XYZ with b=y, a = x, and c>z. 
Prove that Z.OZ.Z. 

The plan is to show that other suppositions lead to an impossibility. 

Proof. Now ZC is either equal to ZZ, less than ZZ, 
or greater than Z.Z. 

If ZC = Z.Z, 

then A ABC is congruent to A XYZ, § 40 

because it then has two sides and the included Z equal respec¬ 
tively to two sides and the included Z. of A XYZ; 

and c = z. § 38 

And if _ ZC<ZZ, 

then c<z. 

Neither conclusion can be true, because c>z. 

.*. ZOZZ. 


§126 

Given 









92 


INEQUALITIES 
Exercises. Review 


BOOK I 


1. The point P within the A ABC is connected with 
A, B, C by the line segments x, y, z as 
shown in this’ figure. Then a + 6 is 
greater than the sum of what two line 
segments? What proposition proves 
your statement? 

A 

2. In Ex. 1, 6 + c is greater than 
what sum, and c + a is greater than what other sum? 

3. In the figure of Ex. 1 write three similar inequalities, 
beginning with x-{-y>c, add the three inequalities, and see 
what interesting result you can find relating to x + y A-^ 
and a + 6 + c. 

4. Draw a figure showing how many exterior angles a 
triangle may have and find their sum in degrees. 

5. In the angles of this figure how does x compare with 2/? 
State the reason. How does y compare with Zy and why ? 
Then how does x compare with z, and 
why? Write a theorem beginning, ”If 
from a point within a triangle lines are 
drawn to any two vertices, the angle 
formed by these lines is greater than... ^ ’. ^ 

6. Draw a rectilinear figure of four sides, and produce 
one of the sides to form an exterior angle. State your 
inference as to the relation of the size of this exterior 
angle to that of any of the nonadjacent interior angles. 
Discuss each possibility in full. 

7. The angles of a certain quadrilateral are so related 
that the second is twice the first, the third three times 
the first, and the fourth four times the first. How many 
degrees are there in each angle ? 







§128 


ATTACKING ORIGINALS 


93 


IV. Attacking Originals 

128. General Suggestions. Various important suggestions 
for attacking those exercises which are often called origi¬ 
nals have already been given in connection with the exer¬ 
cises themselves. These will now be summarized: 

1. Draw the figure carefully, hut do not stop to construct 
it unless there seems to he some special need for doing so. 

A proof is often unnecessarily difficult simply because the figure 
is carelessly or incorrectly drawn. 

2. Draw as general figures as possible. 

For example, if you wish to prove a proposition about any triangle, 
do not take a triangle that is isosceles, right, or equilateral. 

3. After drawing the figure, state precisely what is given 
and precisely what is to he proved. 

Many of the difficulties of geometry come from failing to keep in 
mind precisely what is given and precisely what is to be proved. Draw 
no extra lines unless it is necessary. 

4. Now see if the proof is at once clear. If it is not, say: 
can prove this if I can prove that; I can prove that if I 

can prove.. and so on until you reach a proved propo¬ 
sition. Then reverse your reasoning. 

5. If two line segments are to he proved equal, try to prove 
them corresponding sides of congruent triangles, sides of 
an isosceles triangle, opposite sides of a parallelogram, or 
segments between parallels which cut equal segments from 
another transversal. 

6. If two angles are to he proved equal, try to prove them 
alternate or corresponding angles of parallel lines, corre¬ 
sponding angles of congruent triangles, base angles of an 
isosceles triangle, or opposite angles of a parallelogram. 

7. Try the indirect method (§ 56) as a last resort. 


94 


ATTACKING ORIGINALS 


BOOK I 


129. Synthetic Method. The method of proof in which 
known truths are put together in order to obtain a new 
truth is called the synthetic method. 

This method is used in proving most of the theorems of geometry. 
The proposition usually suggests some propositions already proved, 
and from these we proceed to the proof required. 

130. Analytic Method. The method of attack which asserts 
that a proposition under consideration is true if another 
proposition is true, and so on, step by step, until a known 
truth is reached, is called the analytic method. 

This is the method referred to in the fourth suggestion in § 128. It is 
the one which the student should use if he does not at once see the proof. 

131. Concurrent Lines. If two or more lines pass through 
the same point they are called concurrent lines. 

The word ''concurrent” is from two Latin words meaning "run¬ 
ning together.” Since two lines are generally concurrent, the term is 
commonly used in connection with three or more lines. 

132. Median. A line segment from any vertex of a tri¬ 
angle to the midpoint of the opposite side is called a median 
of the triangle. 

The term is occasionally employed with reference to a trapezoid 
to mean the line segment joining the midpoints of the two nonparallel 
sides, but it is rarely needed for this purpose. 

133. Trisect. To divide any geometric magnitude into 
three equal parts is to trisect it. 

Exercises. Review 

1. How many sides are there in a regular polygon each 
of whose angles is 175° ? 

2. If a side and an angle of one isosceles triangle are 
equal respectively to the corresponding side and angle of 
another isosceles triangle, the triangles are congruent. 


§§ 129-133 


REVIEW EXERCISES 


95 


3. Given the rt. l\ABC with Z.B = 2Z^, with M the mid¬ 
point of the hypotenuse AB, and with MN II to AC, as 
shown in the figure. Give the authority for each of the 
following statements: 

iX)BN=NC, (5)ZA = 30°. 

(2) MN is ± to BC. (6) Z R = 60°. 

(3) MB = MC. (7) MB = BC. 

(4) ZA -h Z^ = 90°. (8) AB^2 BC. 

4. The bisector of an exterior angle of an isosceles tri¬ 
angle, formed by producing one of the equal sides through 
the vertex, is parallel to the base. 

I can prove that AX is II to BC if I can prove x 

that A -and-are equal. I can prove these ^Z \ 

angles equal if I can prove that Z CA Y is twice N. 

Z-of the A ABC. 

5. If the line drawn from the vertex of a triangle to the 
midpoint of the base is equal to half the base, the angle 
at the vertex is a right angle. 

6. If through any point in the bisector of 
an angle a line is drawn to either side of the 
angle parallel to the other side, the triangle 
thus formed is isosceles. 




7. If one of the equal sides of an isosceles triangle is 
produced through the vertex by its own y 

length, the line joining the end of the side 
produced to the nearer end of the base is 
perpendicular to the base. 

can prove that ZYCB is a right angle if I can 

prove that it is equal to the sum of A -and-of 

ABCY. I can prove that it is equal to this sum if I 

can prove that p = Z-and q — Z-Now reverse this reasoning 

and write out the proof in full. 








96 


ATTACKING ORIGINALS 


BOOK I 


8. Through any point P on the line AB an intersecting 
line is drawn, and from any two points on this line equi¬ 
distant from P perpendiculars are drawn to AB or AB 
produced. Prove that these perpendiculars are equal. 

9. The bisectors of two supplementary adjacent angles 

are perpendicular to each other. C 


10. The lines joining the midpoints of 
the sides of a triangle divide the triangle 
into four congruent triangles. 



11. The bisectors of two vertical angles 
are in the same straight line. 

12. The bisectors of the two pairs of 
vertical angles formed by two intersecting 
lines are perpendicular to each other. 



13. If an angle is bisected, and if a line is 
drawn through the vertex perpendicular to 
the bisector, this line forms equal angles with 
the sides of the given angle. 



« 

14. The bisector of the angle at the vertex of an isos¬ 
celes triangle bisects the base and is perpendicular to 
the base. 


15. The perpendicular bisector of the 
base of an isosceles triangle is concurrent 
with the equal sides and bisects the angle 
at the vertex. 



16. If the perpendicular bisector of the base of a triangle 
passes through the vertex, the triangle is isosceles. 

17. Any point on the bisector of the angle at the ver¬ 
tex of an isosceles triangle is equidistant from the ends 
of the base. 

Take any point Pon AM'm the figure of Ex. 15 and show that PB=PC. 







§133 EQUAL LINES 97 

Exercises. Equal Lines 

1. In an isosceles triangle the medians drawn to the 
equal sides are equal. 

2. If the sides AB and AD of a quadri¬ 
lateral ABCD are equal, as shown in this 
figure, and if the diagonal AC bisects the 
angle at A, then BC = DC, 

3. If a line segment is terminated by two parallel lines, 
and if another line segment is drawn through the mid¬ 
point of the first and is terminated by the parallels, the 


second segment is bisected by the first. 

4. In a OABCD the line BQ bisects 

AD, and DP bisects BC, Prove that BQ t 

and DP trisect AC. ^ ^ 

5. If on the base AP of a A ABC any C 

point P is taken, and the lines AP, PP, 


BC, and CA are bisected by W, X, Y, and 
Z respectively, then XY = WZ, ^ ' w' P 'x" ^ 

6. In the square ABCD, if CD is bisected by Q, and if P 
and R are taken on AP so that AP = PP, 
then PQ = RQ, 

7. In this figure, if AC — BC, and if 
= PQ = CP = CP, then PS = QR, 

8. If from the vertex and the mid¬ 

points of the equal sides of an isosceles 
triangle lines are drawn perpendicular to the base, they 
divide the base into four equal parts. ^ ^ 

9. In this figure, if AP is II to DC, if jy ' ' vj 

Z C = Z A and if CP=DQ, then AP=BQ. j \ 

Produce AP and PC to intersect. Then how ^1^1 - 

can it be shown that AD = BC ? 










98 


ATTACKING ORIGINALS 


BOOK I 


Exercises. Equal Angles 


1. If the angles at the vertices of two isosceles tri¬ 
angles coincide, what can be said of the bases ? Prove it. 

2. The bisectors of the equal angles of an isosceles 
triangle form with the base another isosceles triangle. 


3.. In this figure, if AB = AC, and if CQ 
and BR bisect the AYCA and XBA re¬ 
spectively, the triangle formed by pro¬ 
ducing QC and RB is isosceles. 

4. The bisectors of any two angles 
of an equilateral triangle form an angle 
equal to any exterior angle. 



5. In which direction must the side 6 of a AABC be 
produced so as to intersect the bisector of 
the opposite exterior angle ? 

Consider the three cases ZA<ZC, ZA = ZC, 

ZA>ZC. " c B~ 


6. A line drawn parallel to the base of an isosceles 
triangle makes equal angles with the sides or the sides 
produced. 

7. If the bisector of an exterior angle 

of a triangle is parallel to the opposite side, \ 

the triangle is isosceles. 


8. If through the three vertices of an isosceles triangle 
lines are drawn parallel to the opposite sides, they form an 
isosceles triangle. 

9. In the figure here shown, if AD ~ BC, 
and Z A = Z5, then DC is II to AB. 

10. If a line drawn at right angles to 
AB, the base of an isosceles /\ABC, cuts AC at P and BC 
produced at Q, then A PCQ is isosceles. 








§133 


CONGRUENCE 


99 


Exercises. Congruence 

1. If two sides and the included angle of one parallelogram 
are equal respectively to two sides and the included angle of 
anotherj the parallelograms are congruent. 

This proposition is occasionally required in courses of study. In 
proving it the method of § 40 should be used. 

2. If in a A ABC a perpendicular is drawn from B to the 
bisector of Z^, meeting this bisector at X and AC or AC 
produced at U, then 

3. If through any point equidistant from two parallel 
lines two lines are drawn cutting the parallels, they inter¬ 
cept equal segments on these parallels. 

4. If, from the point where the bisector 
of an angle of a triangle meets the opposite 
side, lines are drawn parallel to each of the 
other sides, the segments of these lines cut 
off by the sides are equal. 

5. The diagonals of a square are perpendicular to each 
other and bisect the angles of the square. 

6. If two line segments bisect each other at right angles, 
any point on either segment is equidistant from the ends 
of the other segment. 

7. If either diagonal of a parallelogram bisects one of the 
angles, the sides of the parallelogram are 
all equal. 

8. On the sides of any /\ABC the equi¬ 
lateral ABPCj CQA, ARB are constructed. 

Prove that AP=CR =BQ, 

How can we prove that AABP is congruent to 
ARBC and that A ARC is congruent to A^J5Q? 

Does proving these facts establish the proposition? 








100 


ATTACKING ORIGINALS 


BOOK I 


Exercises. Sums of Angles 


1. An exterior angle of an acute triangle or of a right 
triangle cannot be acute. 

2. If the sum of two angles of a triangle is equal to the 
third angle, the triangle is a right triangle. 

3. If the line joining any vertex of a triangle to the 
midpoint of the opposite side divides the triangle into two 
isosceles triangles, the original triangle is a right triangle. 

4. If the angles at the vertices of two isosceles tri¬ 
angles are supplements one of the other, the base angles 
of the one are complements of those of the other. 


5. If from the ends of the base AB 
of a l\ABC perpendiculars to the other 
two sides are drawn, meeting at P, then 
ZP is the supplement of ZC. 


C 


A^ 


rB 


Here AP is ± to AC and BP is ± to BC. Con¬ 
sider the case in which AP is ± to PC and BP is 


± toAC. 


6. The bisectors of two consecutive angles of a paral¬ 
lelogram are perpendicular to each other. 

7. If two sides of a quadrilateral are parallel, and the 
other two sides are equal but not parallel, the sums of 
the opposite angles are equal. 

8. If the exterior angles at B and C 
of any A APC are bisected by lines meet¬ 
ing at P, then ZP-f J ZA = a rt. Z. 

9. The opposite angles of the quadri¬ 
lateral formed by the bisectors of the interior angles of 
any quadrilateral are supplementary. 

10. The angles of a quadrilateral are x,2x,2 Xy and 3 x. 
How many degrees are there in each angle ? 





§133 


INEQUALITIES 


101 


Exercises. Inequalities 

1. In the AA5C the ZA is bisected by a line meeting 
BC at D, Prove that BA > BD, and that CA > CD. 

While less important than the suggestions given in §128, the fol¬ 
lowing will be found helpful: 

If one angle is to he proved greater than another, try to show that 
it is an exterior angle of a triangle, or an angle opposite the greater 
side of a triangle. 

If one line is to he proved greater than another, try to show that it 
is opposite the greater angle of a triangle. 


2. If AD is the longest side and BC is the shortest side 
of the quadrilateral ABCD, then Z.B>Z.D and ZC>ZA. 

3. If a line is drawn from the vertex A of a 
square ABCD so as to cut CD and to meet BC 
produced in P, then AP>DB. 

4. If the angle between two adjacent sides 
of a parallelogram is increased, the length of 
the sides remaining unchanged, the diagonal from the 
vertex of this angle is diminished. 



5. If a point P is taken within a A ABC such that 
CP = CB, as shown in this figure, then 
AB>AP. 

6. In a quadrilateral ABCD, if AD = BC 
and ZC < ZP, then AC > BD. 

7. In Ex. 6 prove that ZP>ZA. 



8. In a pentagon ABCDE it is given that ZA=ZP<ZC. 
Can you make any inference as to the equality or inequality 
of A C, BD, and BE ? Explain your answer. C 

9. In the A ABC, if AB>AC and if on 
AP and AC respectively BP is taken equal 
toCQ, then BQ>CP. 


A 











102 


ATTACKING ORIGINALS 


BOOK I 


Exercises. Triangles 


1. If two triangles have two sides of one equal respec¬ 
tively to two sides of the other, and the angles opposite 
two equal sides equal, the angles opposite the other two 
equal sides are either equal or supplementary, and if equal 
the two triangles are 
congruent. 

Using superposition, 
as in § 40, and placing 
the corresponding parts 
in the usual way, since 
ZB' = ZB, then B'A' lies along what line? Then A' lies on A or on 
some other point of BA, as D. If A' lies on A, are AA'B'C' and ABC 
congruent ? 

If A' lies on D, are AA'B'C and DBC congruent? 

Since CD = C'A'=CA, what is the relation of ZA to ZCDAl of 
ZCDA to ZBDCl of ZA to ZBDCl 

The triangles are congruent under what conditions with respect to 
A B and B' ? with respect to Z A and A' ? 



C' (f 

t\\ 


2. The midpoint of the hypotenuse of a right triangle 
is equidistant from the three vertices. 

We have to prove that AM=BM, that CM=BM, 
or that BM is half of a line segment that is equal 
to AC. 

This may be proved in several ways. Probably 
the simplest way with this figure is to prove certain 
triangles congruent. Another way would be to adapt the figure to § 83. 



3. If one acute angle of a right triangle is double the 
other, the hypotenuse is double the C 

shorter side. 

This is the familiar 30°-60° right triangle used 
by draftsmen. 

If AM=CM, then AM=:BM=CM, as in 
Ex. 2. The exercise then reduces to proving that ABCM is equilateral 
by proving that p = 2 a = 60° = g. 











§133 


TRIANGLES 103 





\J 


4. A median of a triangle is less than half the sum of 
the two adjacent sides. 

The student should attack this exercise by 
analysis, beginning as follows: 

Given CM, a median of the A-4RC. 

Prove that CM <^{BC+ CA). 

I can show that CM <^{BC+ CA) if I can 
show that 2 CM<BC+ CA. 

This suggests producing CM by its own 
length to P and drawing AP. 

Now CP=2 CM, 

and I can show that 2 CM < RC + CA 

if I can show that CP < BC + CA. 

But CP<AP-\-CA. Post. 3 

Hence CP<BC-{- CA 

if I can show that BC = AP, 

and this is true if AMBC is congruent to A MAP.** 

Now complete the analysis, then reverse the reasoning, and write 
out the proof in full. 


5. The diagonals of a rhombus form four right angles. 

6. The perpendiculars from two opposite vertices of a 
parallelogram drawn to the diagonal determined by the 
other vertices are equal. 

7. From the vertex A of a A AJ5C the line AD is drawn _L 
to BC. Consider the following statements and tell which 
ones are true in general. Then tell what other conditions 
must be given in order that the other statements shall 
be true: 

(1) BD = BC. 

(2) AD<i(AB-\-AC). 

(3) ZADB>ZB. 

(4) Either Z CD A < ZB or CD A <ZC. 




104 


ATTACKING ORIGINALS 


BOOK I 


Exercises. Review 


1. Make a list of the numbered propositions in Book I, 
stating under each the previous propositions upon which 
it depends either directly or indirectly. 


2. Make another list of the numbered propositions, stat¬ 
ing under each the subsequent propositions in Book I which 
depend upon it. 


3. The line joining the midpoints of the nonparallel 
sides of a trapezoid passes through the ^ ^ 

midpoints of the two diagonals. 

How is PQ related to AB and DC? Why ? Since 
PQ bisects AD and BC, how does it divide AC and Ar ————— 

RD? Why? 


B 


4. The lines joining the midpoints of 
the consecutive sides of any quadri¬ 
lateral form a parallelogram. 

How are PQ and SR related to AC? 



5. If the diagonals of a trapezoid are equal, the trapezoid 
is isosceles. P c 

Construct DP and CQ _L to AB. How is AAQC 
related to ABPDl Why? Then how is Z.QAC 
related to ZPBDl Then how is AABD related to ' 

ABAC’i 


^ P Q ^ 


6. If, from the diagonal BD of a square ABCD, BP is cut 
off equal to BC, and PQ is constructed ± 
to BD, meeting the side CD at Q, then 
PD = PQ^QC. 

How is rt. ABQP related to rt. ARQC? Why? 

How many degrees are there in ZPDQ and in 
ZPQD‘i Then how is PD related to PQ? Why? 



7. Study Ex. 6 for the case of BP= i BD, and state and 
prove the resulting proposition. 












§133 


APPLICATIONS 


105 


Exercises. Applications 


1. In order to put in a brace which shall join two 
converging beams and make equal 
angles with them, a carpenter places 
two steel squares as here shown, so 
that OP= OQ, Prove that PQ makes 
equal angles with the beams. 

2. In what other way can you con¬ 
struct the line PQ in Ex. 1 so that it shall make equal 
angles with the beams? 





\ 




2;—— 


"J 


3. Wishing to measure the distance AX in this figure, 

a boy placed a pair of compasses C on top of a post A so 
that one leg was vertical and the ^ 

other pointed to X. He then turned 
the compasses around, keeping the 
angle fixed and the leg on the post 
vertical, and sighted along the other 
leg to Y. He then measured A Y and 

thus found the distance AX. Explain the principle involved. 

4. In the figure, MM' represents a mirror 
and PQ is JL to MM' at P. If a ray of light 
LP from a light L strikes the mirror at P, 
it is reflected to the eye at E in such a way 
that a = 6. The line LL' is ± to MM'y and 
EPL' is a straight line. Prove that x = y 
and that y = Zy and explain why the light 
appears to be at the same distance behind the mirror, at L', 
that it really is in front of it, at L. 



5. This figure represents a 'parallel ruler whioh is used for 
drawing parallel lines. Explain how 
it may be used, and state the theorem 
upon which its principle depends. 
















106 


ATTACKING ORIGINALS 


BOOK I 


6. It is proved in physics that two forces acting on 
an object O have the same effect as a single force known 
as their resultant. If, to scale, we let 
OX represent a force of 300 lb. pulling 
in the direction OX, and OK a force of 
1501b. pulling in the direction OF, the 
resultant is represented by the diagonal OR of the O OXRY. 
By measuring OR and XXOR we can find the magnitude 
and the direction of the resultant. Using a protractor and 
ruler, find the magnitude and the direction of the resultant 
of OX and OF in the above case. 

7. Two forces at right angles to each other are exerted 
upon an object. One force is 500 lb. to the right and the 
other is 800 lb. upward. Find the resultant as in Ex. 6. 

8. Explain geometrically why this ^ 

telephone extends horizontally when it || 

is pulled out. State each proposition 
involved in the answer. What would I /|| 

be the effect on the direction if each bar extending || 
from the top downward to the right were longer than Ml 
each bar extending downward to the left ? ^ 

9. To ascertain the height of a tree or of the school 
building, fold a piece of paper so 
as to make an angle of 45°, or take 
a draftsman’s 45-degree triangle; 
then walk back from the tree until 
the top is.seen at an angle of 45° 
with the ground, being careful to 
hold the base of the triangle level. In 
the figure prove that AB=ACy and 
hence that CX— AB-\-BYy where BY is the height of the 
observer’s eye above the ground. 










§133 


APPLICATIONS 


107 


10. This figure represents four hinged rods with AB=DC 

and AD—BC, As the angles change, do6s the figure con¬ 
tinue to be a parallelogram? Upon n_ n 

what theorem does this depend? ^ 

11. In the figure of Ex. 10, if Z A is 
125°, how large are A By C, and D^. ^ 




12. Explain how this instrument, in which the joint X 
can be moved along the rod OX, 
is used to bisect an angle with 
the sides of which the arms OA 
and OB can be made to coin¬ 
cide. State the propositions on 
which the explanation depends. 

On account of the joints and other 
mechanical features of such an instrument this method of bisection 
is not so nearly accurate as the construction given in § 103. 



13. The simple bridge construction here shown is occa¬ 
sionally used. The beams PA and P'A rest on the per¬ 
pendicular support OA in the center 
of the bridge. The rods OP and OP' 
are fastened at O, P, and P'. Show 
by means of the congruence of cer¬ 
tain triangles that the point O always remains directly 
beneath A. Why will the bridge support a weight? 



14. In laying out a tennis court it is desired to run a line 
through a point P II to AP. This is a convenient method: 
Stretch a tape from P to any point Q P . \S 

on AP; then with Q as center swing / 

the tape to cut AP at R ; with P and R 

as centers and the same radius as before ^—l -1- ^ 

mark arcs intersecting at S, and draw ^ ^ 

a line through P and P. Prove that PS is the line required. 













108 


ATTACKING ORIGINALS 


BOOK I 




15. In laying out a tennis court, another way of running 
a line through a point P II to AB is as follows: A line is 
drawn from P to Q, any point on AB, 
and the midpoint M of PQ is found 
with a tape. Then from another point 

X on AB the tape is stretched through ^_ 

M, and a point Y is found such that ^ ^ 

MY = XM. Then CD, drawn through Y and P, is II to AB. 


9M 


16. A board 8 in. wide is to be sawed into five strips of 
equal width. In order to draw the lines for sawing, a 
carpenter lays his steel square as 
here shown, placing the corner on 
one edge and the 10-inch mark 
on the other, and marks the board 
at the divisions 2, 4, 6, 8 on the 
square. He then moves the square 
along the board and repeats the 
process. Prove that lines drawn through the corresponding 
marks satisfy the requirements. 



17. A dentist's working table, in which bar x is fastened 
to bar y at right angles and table t is 
fixed parallel to bar x, is attached to 
a vertical wall W as shown in this 
figure. State in full the geometric proof 
that table t is always horizontal. 



18. This figure represents six hinged rods in which all 
the angles are right angles and P, Q, 

R, S bisect AB, BC, CD, DA respectively. 

Prove that the figure can be pulled into 
different shapes, the angles then ceas¬ 
ing to be right angles, but that all the "" ^ ^ 

quadrilaterals will still continue to remain parallelograms. 



















BOOK II 


THE CIRCLE 

I. Fundamental Theorems 

134. Properties of a Circle. From the definitions in § 22 
and from a study of the figure we see that a circle has 
certain properties, among which are the following: 

1. All radii of the same circle or of equal circles are equal, 

2. All circles with equal radii are equal. 

3. All diameters of the same circle or of equal circles are 
equal. 

A. If a straight line intersects a circle in one point, it 
intersects it in two points and only two. 

5. If two circles intersect in one point, they intersect in 
two points and only two. 

6. A point is within, on, or outside a circle according as 
its distance from the center is less than, equal to, or greater 
than a radius. 

7. A diameter bisects the circle and the surface inclosed, 
and conversely. ■ 

These statements may be taken as postulates and referred to as 
properties of a circle, although ^they are capable of proof. 

135. Central Angle. If the vertex of an angle is at the 
center of a circle and the sides are radii of the circle, the 
angle is called a central angle. 

An angle is said to intercept any arc cut off by its sides, 
and the arc is said to subtend the angle. Preferably, we 
speak of the arc as having a central angle, and conversely. 

109 


110 


FUNDAMENTAL THEOREMS 


BOOK II 


Proposition 1. Equal Angles have Equal Arcs 

136. Theorem. If two central angles of the same circle 
or of equal circles are equaly the angles have equal arcs; 
and if two central angles are unequal, the greater angle 
has the greater arc. 



Given the equal © 0 and O' with central ZA05= central 
Z.A'0'B^ and with central ZAOC> central ZA'0'5'. 

Prove that arc AB = arc A'B' and that arc AC > arc A'B', 
The best plan is to place one figure on the other. 

Proof. Place O O on O O' so that /LAOB coincides with 


its equal, Z.A'0'B', Post. 5 

In the case of the same O simply swing one Z about O. 

Then A lies on A' and B on B\ § 134,1 

.■. arc AB coincides with arc A-B\ § 21 

because all points of each are equidistant from O'. 

Since /-AOC>/LA'O'B' ^xi6. Z.AOB =/.A'0'B\ Given 
we have ZAOOZAOB, Ax. 5 

.\OC lies outside ZAOB, § 6 

and hence arc AC > arc AB. Ax. 10 

But arc AB= arc A 'B Proved 

and hence arc AC> arc A'B Ax. 5 










§§ 136-138 


CENTRAL ANGLES 


111 


Proposition 2. Equal Arcs have Equal Angles 

137. Theorem. If two arcs of the same circle or of 
equal circles are equal, the arcs have equal central 
angles; and if two minor arcs are unequal, the greater 
arc has the greater central angle. 

Given the equal © O and O' with arc AB = sltcA'B' , minor 
arc AC > minor arc A'B', and the central AAOB, A'O’B', AOC, 

Prove that AAOB=Z.A'O'B' and that AAOOAA'O'B'. 

The best plan, as in § 136, is that of superposition. 

Proof. Using the figure of § 136, place O O on O O' so 
that OA shall lie on its equal, O'A', and the arc AB on its 


equal, the arc A'B'. 

Post. 5 

Then 

OB coincides with O'B'. 

Post. 1 


ZAOB^ZA'O'B', 

§10 

thus proving the first part of the theorem. 


Since 

arc AC> arc A'B’, 

Given 

we have 

arc AO arc AB, 

Ax. 5 


because arc A'B' is given equal to arc AB; ' 


and hence 

OB lies within A AOC, 


because otherwise we could not have arc AO arc AB. 


ZAOOZAOB, 

Ax. 10 

and hence 

ZAOOZA'O’B', 

Ax. 5 


thus proving the second part of the theorem. 

This proposition is the converse (§ 70) of the one in § 136. 

138. Chord. A straight line that has its ends 
on a circle is called a chord of the circle. 

A chord is said to subtend the arcs that it cuts from 
a circle, but it is more simple to speak of the chord of 
the arc. Unless the contrary is stated, the chord is to 
be considered as belonging to the minor arc. 




112 


FUNDAMENTAL THEOREMS 


BOOK II 


Proposition 3. Equal Arcs have Equal Chords 

139. Theorem. If two arcs of the same circle or of 
equal circles are equal, the arcs have equal chords; and 
if two minor arcs are unequal, the greater arc has the 
greater chord. 



Given the equal © 0 and O' with arc AB — arc A'B' and with 
minor arc AX'> minor arc A'B'. 

Prove that chord AB = chord A'B' 
and that chord AX > chord A’B', 

The plan is to show that two A are congruent and that the greater 


chord is opposite the greater Z of a A. 

Proof. Draw radii to A, B, X, A', B', Post. 1 

Since 0A = 0'A' and 0B=0'B', § 134,1 

and ZA0B = ^A'0'B', § 137 

we see that /AOAB is congruent to /SO’A'B', § 40 

and hence chord AR= chord A'R', §38 

thus proving the first part of the theorem. 

Then in AO AX and O'A'B' we have 

0A=0'A' and OX= O 'B', § 134,1 

while ZA0X>Za'0'B', §137 

.*. chord AX > chord A'B', § 126 

thus proving the second part of the theorem. 








§§ 139,140 


CHORDS AND ARCS 


113 


Proposition 4. Equal Chords have Equal Arcs 

140. Theorem. If two chords of the same circle or of 
equal circles are equal, the chords have equal arcs; and 
if two chords are unequal, the greater chord has the 
greater minor arc. 



Given the equal © 0 and O' with chord AB = chord A'B' and 
with chord AX'> chord A'B'. 

Prove that arc AB = arc A'B' 

and that arc AX > arc A'B\ 

The plan is to show that the equal chords have equal central A and 
that the centralZof the greater chord is opposite the greater side of a A. 


Proof. Draw radii to A, B, X, A', B'. Post. 1 

Since 0A = 0'A'and 0B=0’B', § 134,1 

and chord AB= chord A'B', Given 

we see that A GAB is congruent to AO'A § 47 
AA0B = Z.A'0'B\ §38 

and hence arc AB — arc A'B\ §136 

Then in AO AX and O'A'B' we have 

0A=^0'A'and 0X=0'B', §134,1 

while chord AX > chord A'B', Given 

ZA0X>ZA'0'B', §127 

arc AX > arc A§136 


and 









114 


FUNDAMENTAL THEOREMS 


BOOK II 


Proposition 5. Diameter Perpendicular to a Chord 

141. Theorem. If a diameter is perpendicular to a 
chord, it bisects the chord and its two arcs. 



Given the O 0 with a diameter PQ _L to a chord AB at M, 


Prove that 

AM = BM, 


that 

arc AQ= arc BQ, 


and that 

arc AP — arc BP, 


The plan is to prove first that two A are congruent. 


Proof. Draw 

radii to A and B, 

Post. 1 

Since PQ is given _L to AB, A AMO and BMO are rt. A. § 20 

Then since 

OM=OM, 

Iden. 

and 

OA = OB, 

§ 134,1 

A AMO is congruent to A BMO, 

§71 


AM = BM. 

§38 

Likewise, 

ZAOQ=^ZBOQ, 

§38 

and 

ZPOA=^ZPOB; 

Post. 9 

hence arcAQ = 

arc BQ, and arc AP = arc BP. 

§136 


142. Corollary. If a diameter bisects a chord which is not 
itself a diameter, it is perpendicular to the chord. 

Show that § 47 applies. 









§§141-143 CHORDS AND ARCS 115 

143. Corollary. The perpendicular bisector of a chord 
passes through the center of the circle and bisects the arcs 
of the chord. 

How many _L bisectors of the chord are possible ? Then with what 
line must the ± bisector coincide (§ 141) ? Complete the proof. 


Exercises. Chords and Arcs 


1. The greater of two unequal major arcs has the 
shorter chord. 

Prove that this follows from § 139. 


2. The greater of two unequal chords has the shorter 
major arc. 


Prove that this follows from § 140. 

3. If AABC is an equilateral triangle, find 
the number of degrees in the central AAOB, 
BOC, COA and in the arcs AB, BCy CA, State 
the reason in each case. 



4. If a radius bisects an arc it bisects the 
chord of that arc. 

5. If a radius bisects a chord which is not 
a diameter, it bisects its central angle. 



6. If a diameter bisects a chord which is not itself a 
diameter it bisects the two arcs of the chord. 


7. The line bisecting the two arcs which have the same 
chord is the perpendicular bisector of the chord. 

8. If a wheel has eight spokes, spaced 
equally, how many degrees are there in each 
of the eight small arcs thus formed? State 
the reason involved in the answer. 

9. The chord of half an arc is greater than half the 
chord of the whole arc. 







116 


FUNDAMENTAL THEOREMS 


BOOK II 


144. Tangent. An unlimited straight line which touches 
a circle at only one point is said to be tangent to the circle. 
Such a line is called a tangent to the circle. 

For example, in this figure t is tangent to the 
circle at the point P. 

The word ''tangent” is from the Latin word 
tangere, to touch. Hence we may say that a line 
touches a circle instead of saying that it is tangent 
to the circle. If a line is tangent to a circle, the circle is also said to 
be tangent to the line. 

The point at which a tangent touches the circle is called 
the point of tangency or point of contact. 

Although a tangent is unlimited in length, when we 
speak of a tangent from an external point 
to a circle we mean the segment between 
the point and the circle. 

For example, the tangent from P to the circle 
here shown is the segment PT. 

145. Tangent Circles. Two circles which are both tangent 
to the same line at the same point are called tangent circles. 

Circles are said to be tangent externally or tangent 
internally according as they lie on opposite sides or on the 
same side of the tangent line. 

For example, in the first of these 
figures the circles are tangent exter¬ 
nally, and in the second figure they are 
tangent internally. 

The point of contact of two tangent circles with the 
tangent line is called the point of contact or point of tan¬ 
gency of the circles. 

In the first of the two figures just above, the line t is 
called a common internal tangent, and in the second a com¬ 
mon external tangent. 







§§ 144-148 


TANGENTS 


117 


Proposition 6. Condition of Tangency 

146. Theorem. If a line is perpendicular to a radius 
at its end on the circle, the line is tangent to the circle. 



Given the G O with the line XY _L to the radius OP at P. 


Prove that XY is tangent to the O. 

The plan is to show that all points on XY except P are outside the O. 


Proof. Let A be any point on AF except P,. and draw OA, 
Then OA > OP, § 122 

and hence A is outside the O. § 134, 6 


Then every point on XY except P lies outside the O, 
and hence XY is tangent to the O. § 144 

147. Corollary. If a line is tangent to a circle, it is perpen¬ 
dicular to the radius drawn to the point of contact. 

Since every point on XY except P is outside the O, then OP is the 
shortest line segment from O to XY. Hence Z OPX is a rt. Z (§ 123). 


148. Corollary. If a line is perpendicular to a tangent at 
the point of contact, it passes through the 
center of the circle. 

A radius OT is _L to a tangent at T’(§ 147). If a ±, 
say TS, constructed to XY at T, did not coincide with 
this radius, we should have two J§ to XY at the same 
point T, which is impossible (Post. 10). 











118 


FUNDAMENTAL THEOREMS 


BOOK II 


Proposition 7. Lengths of Tangents 

149. Theorem. The tangents to a circle from an exter¬ 
nal point are equal and make equal angles with the line 
joining the point to the center. 



Given PA and PR, tangents from an external point P to the 


O 0, and also given PO, the line joining P to O. 

Prove that PA=PB 

and that Z. OP A = Z OPB. 

The plan is to prove that two A are congruent. 

Proof. Draw the radii OA, OB, Post. 1 

Now PA is _L to OAy 

and PB is ± to OR, § 147 

because if a line is tangent to a O, it is _L to the radius 
drawn to the point of contact. 

.*. AAPO and BPO are rt. A. § 20 

Then in AAPO and BPO we have 

PO^PO, Iden. 

and OA = OB, §134,1 

.*. AAPO is congruent to A BPO, § 71 

Hence PA—PB 

and ZOPA=ZOPB, §38 








§149 


REVIEW EXERCISES 


119 


Exercises. Review 




1. A perpendicular from the center of a circle to a tangent 

passes through the point of contact, ^ 

2. In this circle the chords AM and BM are 
equal. Prove that M bisects the arc AB and 
that the radius OM bisects the chord AB, 

3. If P is a point on a circle such that it is 
equidistant from two radii OA and OP, then P 
bisects the arc AB, 

4. If five points A, P, C, P, E are so placed 

on a circle that AB, BC, CD, DE are equal 
chords, then AC, BD, CE are equal chords, and 
AD and BE are also equal chords. q 

5. If tangents to a circle at the points 
A, B, C meet in P and Q, as here shown, C 
then AP-\-QC = PQ, 

Apply § 149 twice. 

6. If a quadrilateral has each side 
tangent to a circle, the sum of one pair 
of opposite sides equals the sum of the 
other pair. 

In this figure show that SP + QR = PQ + RS. 

Apply § 149 four times. 

7. The hexagon here shown has each 
side tangent to the circle. Prove that 

AB+CD + EF = BC + EE-\-FA, 

8. If a quadrilateral has each side tan¬ 
gent to a circle and if the vertices are 
joined to the center, the sum of the angles at the center 
opposite any two opposite sides is equal to a straight angle. 




D 









120 


FUNDAMENTAL THEOREMS 


BOOK II 


Proposition 8. Equal Chords 

150. Theorem. Equal chords of the same circle or of 
equal circles are equidistant from the center. 



Given the O 0 with chord AB = chord CD. 

Prove that AB and CD are equidistant from O. 

The plan is to prove that two A are congruent. 

Proof. Let OP be _L to AB and let 00 be _L to CD. § 116 


Draw OA and OD. Post. 1 

Then A CAP and ODQ are rt. A § 20 

with AP= iAB and DQ = iCD. § 141 

Since AB=CD, Given 

then AP=DQ. Ax. 4 

Also, OA = OD. §134,1 

.*. A CAP is congruent to A ODQ, § 71 

and hence OP=OQ; §38 

that is, AB and CD are equidistant from 0. § 73 


Although equal © are mentioned here and in several subsequent 
theorems, it is evidently necessary to consider only a single O. 

151. Corollary. Chords that are equidistant from the center 
of a circle are equal. 

If OP = OQ, then, since OA = OD, we have two congruent rt. A (§ 71). 
Then AP=DQ (§ 38), and hence AB = CD (§ 141 and Ax. 3). 







§§ 150-152 EQUAL AND UNEQUAL CHORDS 121 

Proposition 9. Unequal Chords 

152. Theorem. The less of two chords of the same circle 
or of equal circles is more remote from the center. 



Given the O 0 with chord CD < chord AB. 

Prove that CD is more remote from 0 than AB. 

In the figure the plan is to prove that OR > OP and that OS > OR. 


Proof. Since CD<AB, Given 

we have arc CD < arc AB. § 140 

Suppose that arc AX = arc CD, and draw AX. Post. 1 

Then AX=CD. §139 

Let the _k from O upon AB, CD, AX be OP, OQ, OS 
respectively, and designate the intersection of OS and 
AB as i?. § 116 

Then OS = OQ. §150 

Equal chords • • • are equidistant from the center. 

Also, OR>OP. §122 

The J_ is the shortest line • • • from a given external point. 

But OS>OR, Ax. 10 

so that OS>OP, Ax. 9 

and hence OQ>OP; Ax. 5 

that is, CD is more remote from 0 than AB. § 73 








122 


FUNDAMENTAL THEOREMS 


BOOK II 


Proposition 10. Chords Unequally Distant 

153. Theorem. If two chords of the same circle or of 
equal circles are unequally distant from the center, the 
chord more remote is the shorter. 



Given the O 0 with two chords, AB and CD, such that CD 
is more remote from 0. 

Prove that CD<AB. 

The plan is to show that any other possibility violates § 152 or § 150. 


Proof. Now CD must be greater than AB, equal to AB, 
or less than AB, 


If 

CD>AB, 


then 

AB is more remote from 0. 

§152 

If 

ft? 

II 

8 


then 

AB and CD are equidistant from 0. 

§150 

But 

CD is more remote from 0, 

Given 


.*. CD<AB, 

This proposition is the converse of the one in § 152. The student has 
probably concluded that where we have only three possible conditions, 
as we do in §§ 150 and 152, the converses are always true. 

154. Corollary. A diameter of a circle is greater than any 
other chord. 

For no other chord can be as near the center. 








§§ 153-155 


CHORDS AND TANGENTS 


123 


Proposition 11. Parallels and Arcs 

155. Theorem. If two parallel lines intersect a circle 
or are tangent to it, they intercept equal arcs. 



1. Given a O with AR, a tangent atP, II to a chord CD, 
Prove that arc CP = arc DP, 

The plan is to show first that certain arcs are equal by § 141. 

Proof. Let PP' be JL to AB at P, Post. 10 

Then PP' is a diameter (§ 148), and is also J_ to CD (§ 63). 


Hence arc CP = arc DP, § 141 

2. Given a O with AB and CD, two II chords. 

Prove that arc AC = arc BD, 

Proof. Let XY, a tangent at Q, be II to CD. § 52 

Then XFislltoAP. §58 

. *. arc AQ = arc BQ, and arc CQ = arc DQ. Case 1 
Hence arc AC = arc BD, Ax. 2 

3. Given a O with ARB and CSD, two II tangents. 

Prove that arc RXS = arc R YS, 

Proof. Let chord XY be II to AB. § 52 


Now complete the proof by § 58, Case 1 (above), and Ax. 1. 

















124 FUNDAMENTAL THEOREMS book ii 


156. Inscribed and Circumscribed Polygons. If the sides of 
a polygon are all chords of a circle, the polygon is said to 
be inscribed in the circle; if the sides are all tangents, the 
polygon is said to be circumscribed about the circle. 



Inscribed Polygon Circumscribed Polygon 


The circle is said to be circumscribed about the inscribed 
polygon and to be inscribed in the circumscribed polygon. 

157. Concentric Circles.. Two circles which have the same 
center are said to be concentric. 

For example, the two circles in the first figure below are concentric. 

158. Line of Centers. The line determined by the centers 
of two circles is called the line of centers. 



centers of the © O and O'. 


159. Secant. A straight line which intersects a circle is 
called a secant. 

In this figure the line AB is a secant. 

It is readily inferred from the figure that a 
secant can intersect a circle in only two points, 
and the student should notice that this is further 
evidence of the truth of the statement given in 
§ 134, 4. This property of a circle will be stated and proved later as a 
corollary (§ 192); but until then § 134,4, may be assumed as a postulate. 




§§ 156-159 


REVIEW EXERCISES 


125 


Exercises. Review 


1. If an equilateral triangle and a square are inscribed 
in a circle, the sides of the square are more remote from 
the center than the sides of the triangle. ^ 

2. The shortest chord that can be drawn 
through a given point within a circle is 
the one which is perpendicular to the diam¬ 
eter through the point. 

Show that any other chord CD, through P, is nearer O than is AB. 



3. In this figure, if the diameter CD bisects 
the arc AB, then /LCBA=^CAB. 

What kind of triangle is A^BC? 


D 


4. In two concentric circles it is given 

that MN is a diameter of the larger circle 
and PQ an intersecting diameter of the 
smaller circle. Prove that P, M, Q, and N are the vertices 
of a parallelogram. q 

5. In this figure arc AB'> arc BC and OP 
and OQ are perpendiculars from the center 
upon AB and BC respectively. Prove that A 



ZQPOZPQO, 

6. Three equal chords AB, BC, CD are 
taken end to end, and the radii OA, OB, OC, 
OD are drawn. Prove that A.AOC — Z.BOD 
and state any other pairs of equal angles. 

7. All equal chords of a circle are tangent 



to a concentric circle. 

8. If a number of equal chords are drawn 
in this circle, the figure gives the impres¬ 
sion of a second circle inside the first and 
concentric with it. Explain the reason. 









126 


FUNDAMENTAL THEOREMS 


BOOK II 


9. If two circles are concentric, chords of the larger 
circle that are tangents to the smaller are equal. 

10. Two equal circles cut two equal chords from a secant 
drawn parallel to the line of centers. 

11. If two intersecting chords make equal 
angles with the diameter through their point 
of intersection, the chords are equal. 

12. If two equal chords intersect, the seg¬ 
ments of one are equal respectively to the 
segments of the other. 

13. In this figure, XY is a diameter ± to 
the II chords AB and CD, arc BD = 40°, and 
arc DX = 50°. How many degrees are there 
in the arcs AC, CA, AY, and YBl 

14. In this figure, XY is tangent to the 
circle at P, the chord AB is ± to the diam¬ 
eter PQ, and the arc AQ=125°. How many 
degrees are there in arc PP? 

15. If from any number of points on the 
larger of two concentric circles tangents 
are drawn to the smaller circle, these tangents are equal. 

16. In this figure, AP and CQ are parallel 
tangents which are cut by a third tangent 
QP. If O is the center of the circle, prove 
that ZPOQ = 90°. 

What is the relation of the ^QPA and PQC? 

How do OP and OQ divide these angles ? Now consider the angles 
of the APQO. 




17. If AP is a diameter of a circle with center O, and if 
PC is any chord from P, then a radius OP which is II to PC 
and lies within Z.CBA bisects the arc CA. 











§§ 160,161 


TANGENT CIRCLES 


127 


Proposition 12. Tangent Circles 

160. Theorem. If two circles are tangent to each other, 
the line of centers passes through the point of contact. 



Given the © 0 and O' tangent at P and the line of centers 00', 

Prove that P is on the line of centers. 

The plan is to show that 00' is ± to the common tangent, which is 
left for the student to prove. Although §§ 160 and 161 are not required 
in standard courses, they have many interesting applications. 


Proposition 13. Line of Centers 

161. Theorem. If two circles intersect^ the line of cen¬ 
ters is the perpendicular bisector of their common chord. 



The proof of this proposition is left for the student. 


















128 


FUNDAMENTAL THEOREMS 


BOOK II 


Exercises. Review 

Describe the relative position of two circles if the line 
segment joining the centers is related to the radii as stated 
in Exs. 1-3, and illustrate each case by a figure: 

1. The segment is greater than the sum of the radii. 

2. The segment is equal to the sum of the radii. 

3. The segment is less than the sum but greater than 
the difference between the radii. 

4. If two circles are tangent externally, the tangents 
to them from any point of the common internal tangent 
are equal. 

5. If two circles tangent externally are tangent to a line 
AB at A and B, their common internal tangent bisects AB, 

6. The line drawn from the center of a circle to the 
point of intersection of two tangents is the perpendicular 
bisector of the chord which joins the points of contact. 

7. The diameters of two circles are 8.15 in. and 6.22 in. 
respectively. Find the distance between the centers of the 
circles if they are tangent externally. Find the distance 
between the centers of the circles if they are tangent 
internally. 

8. Three circles of diameters 2.4 in., 1.8 in., and 2.1 in. 
are tangent externally, each to the other two. Find the 
perimeter of the triangle formed by joining the centers. 

9. If two circles tangent externally at P are tangent to 
a line ARat A and B, then ABPA — 90°. 

10. If two radii of a circle, at right angles to each 
other, when produced are cut at A and 5 by a tangent to 
the circle, the other tangents from A and B are parallel 
to each other. 


§§ 162-165 


MEASURES 


129 


162. Measure. The number of times a quantity contains 
a unit of the same kind is called the numerical measure of 
the quantity, or simply its measure. 

For example, the numerical measure of the length of a room in feet 
is the number of times the length contains the unit of length, 1 ft. 

163. Commensurable Magnitudes. Two magnitudes of the 
same kind which can both be expressed as integers in terms 
of a common unit are called commensurable magnitudes. 

For example, 2J: sq, ft. and 3 sq. ft. are commensurable, for 1 sq. ft. 
is contained 9 times in the first and 12 times in the second. In this case 
the common unit taken was ^ sq. ft; but any unit fraction, say of 
this unit is also a common unit. 

Any common unit used in measuring two or more com¬ 
mensurable magnitudes is called a common measure of the 
magnitudes. Each of the magnitudes is called a multiple 
of any common measure. 

164. Incommensurable Magnitudes. Two magnitudes of 
the same kind which cannot both be expressed in integers 
in terms of a common unit are called incommensurable 
magnitudes. 

The diagonal and the side of a square are, as we shall later prove, 
incommensurable lines. We also have incommensurable numbers such 
as 2 and Vs, for there is no number which is contained in both of these 
numbers without a remainder. 

165. Ratio. The quotient of the numerical measures of 
two magnitudes expressed in terms of a common unit is 
called the ratio of the magnitudes. 

Thus, if a room is 20 ft. by 35 ft., the ratio of the width to the length 
is 20 ft. - 4 - 35 ft., or | %, which reduces to |. Here the common unit is 1 ft. 

The ratio of a to 6 is written or a: 6, as in arithmetic and algebra. 

0 

While we shall ordinarily use the first form, the form a: 6 is some¬ 
times convenient. Thus the ratio of 20° to 30°, which is §g, or §, may 
also be written 2:3. 


130 


FUNDAMENTAL THEOREMS 


BOOK II 


166. Incommensurable Ratio. The ratio of two incommen¬ 
surable magnitudes is called an incommensurable ratio. 

Although the exact value of such a ratio cannot be 
expressed by an integer, a common fraction, or a decimal 
fraction of a limited number of places, it may be expressed 
approximately. For example, \/2 = 1.41421356 • • •, which 
is greater than 1.414213 but less than 1.414214, and there¬ 
fore differs from either by less than 0.000001. 

By carrying the decimal further an approximate value 
may be found that will differ from the ratio by less than a 
billionth, a trillionth, or any, other assigned value. That is, 
for practical purposes all ratios are commensurable. 

For the present we shall consider only the ratios of commensur¬ 
able geometric magnitudes. For the incommensurable cases see the 
optional work in §§ 515-517. 

167. Segment. A portion of a plane bounded by an arc 
of a circle and its chord is called a segment of the circle. 

In the figure of § 168 the part above the chord AB is a minor seg¬ 
ment of the circle, and the part below is a major segment. 

168. Inscribed Angle. An angle with its vertex on a circle 
and with chords for its arms is called an 
inscribed angle. 

In the figure here shown, i is an inscribed angle. 

An angle is said to be inscribed in a 
segment if its vertex is on the arc of the 
segment and its arms pass through the ends of the arc. 

In the figure above, i is inscribed in the minor segment. 

169. Sector. A portion of a plane bounded by two radii 
and the arc of the circle which is cut off 
by the radii is called a sector. 

In this figure the shaded portion AOB is a sector of 
the circle. If AB is a quarter of the circle, it and its 
sector are each called a quadrant. 





§§i6e-i7i 


CENTRAL ANGLES 


131 


Proposition 14. Central Angles 

170. Theorem. Two central angles of the same circle 
or of equal circles have the same ratio as their arcs. 



Given the O O with the central AAOB and ^OC. 


Prove that 


AAOB __ arc AB 
A BOG arcBC 


The plan is to assume that the A and their arcs are commensurable. 


Proof. Suppose that some AAOM is contained 3 times 
m AAOB and 5 times in A BOG, § 163 


AAOB AAOM _S 
ABOG 5: AAOM b 


Construct A equal to AAOM as shown. 
Then . the arcs of these A are equal, 


and 

Hence 


arc AB _ 3 • arc AM _ 3 
dire BO 5* arc AM 5 
AAOB _ SLYC AB 
ABOG Sire BO' 


§165 

§106 

§136 

§165 
Ax. 5 


The proof is the same if any other numbers are used. 

171. Angle and Arc Measure. Since the central angles 
contain the same number of units as their arcs, the 
angles and their arcs have the same numerical measure. 
Briefly stated, a central angle is measured by its arc. 

















132 


FUNDAMENTAL THEOREMS 


BOOK ir 


Proposition 15. Inscribed Angle 

172. Theorem. An inscribed angle is measured by half 
its intercepted arc. 



Given the O 0 with the inscribed intercepting arc AC. 


Prove that A Bis measured by i arc AC. 

In the first figure the plan is to show that ZB = ^ ZAOC. 

Proof. 1. If O is on AB, draw OC. Post. 1 

Then since OC = OB, § 134,1 

we have zlB = ZlC. ' §42 

Then since ZlB-\-ZlC = Z.AOC, §66 

we have 2zlB = ZlAOC; Ax. 5 

whence ZB=^ ZAOC. Ax. 4 

Since Z A OC is measured by arc AC, § 171 

then ^ Z AOC is measured by J arc AC. Ax. 4 

.‘. ZR is measured by I arc AC. Ax. 5 

2. If O lies within ZB, draw the diameter BD. Post. 1 

Then ZABD is measured by I arc AD, 

and ZDBC is measured by I arc DC. Case 1 

.*. ZABD-\-ZDBC is measured by ^ arc (AD + DC); Ax. 1 
that is, ZD is measured by o arc AC. Ax. 10 










§§ 172-176 


INSCRIBED ANGLE 


133 


3. If O lies outside /LB, draw the diameter BD. Post. 1 

Then Z.DBC is measured by ^ arc DC, 
and is measured by 2 arc DA. Case 1 

.'. /LDBC — /LDBA is measured by 2 arc {DC — DA)', Ax. 2 
that is, ZD is measured by I arc AC. Ax. 5 

It should be observed that the expression ''ZD is measured by 
I arc AC” is only a shortened form for the expression " The measure 
of ZD = the measure of ^ arc AC. ” Furthermore, " measure of ^ arc 
AC’’ is equivalent to " | the measure of arc AC,” and hence the first 
expression is really an equation and the axioms of equations may be 
applied. 

173. Corollary. An angle inscribed in a semicircle is a 

right angle. 4^—^ 

Show that Z A is half the central st. ZBOC. 

Instead of proving the corollary in this way, it may --—-^C 

be shown that Z A is measured by half of what arc • V ^ / 

It is then what kind of Z ? - 

174. Corollary. An angle inscribed in a segment greater 
than a semicircle is an acute angle, and an angle inscribed 
in a segment less than a semicircle is an 
obtuse angle. 

In giving the proof draw the radii OC, OD. Then 
show that ZA is half the Z COD. Finally, show that 
ZB is half the reflex ZDOC {% 16), which is greater 
than a st. Z. 

175. Corollary. Angles inscribed in the 
same segment or in equal segments are equal. 

Show that each of the ZA, B, C is half the same 
central Z. 

176. Corollary. If a quadrilateral is in¬ 
scribed in a circle, the opposite angles are supplementary. 

Consider Z A and D in the figure of § 174. Their sum is measured by 
half the sum of what two arcs? Give the proof in full. 






134 


FUNDAMENTAL THEOREMS 


BOOK ir 


Exercises. Review 


1. The shorter segment of the diameter through a given 
point within a circle is the shortest line 
that can be drawn from that point to the 
circle. 

Let P be the given point. Prove that PA is shorter 
than any other line PX from P to the circle. 



2. The longer segment of the diameter through a given 
point within a circle is the longest line that can be drawn 
from that point to the circle. 

3. The diameter of the circle in¬ 
scribed in a right triangle is equal to 
the difference between the hypotenuse 
and the sum of the other two sides 



4. A line from a given point outside a circle passing 
through the center contains the short¬ 
est line segment that can be drawn 
from that point to the circle. 

Let P be the point and O the center. How 
does PC + CO compare with PO ? 



5. Through one of the points of inter¬ 
section of two circles a diameter of each 
circle is drawn. Prove that the line which 
joins the ends of the diameters passes 
through the other point of intersection. 



6. The captain of a ship sailing along 
the course AB is informed by his chart 
that the horizontal danger angle (/IL'XL) 
for a reef lying off the coast near two 
lighthouses L and V is 30°. How can the 
captain avoid the reef and where should 
he change his course ? 








§177 


TANGENT AND CHORD 


135 


Proposition 16. Tangent and Chord 

177. Theorem. An angle formed by a tangent and a 
chord drawn from the point of contact is measured by 
half its intercepted arc. 



Given a O with the tangent MN, through P, and the chord 
PQ making the x. 

Prove that x is measured by arc PSQ. 

In the figure the plan is to show that x = y, that the arcs QSP and 
PR are equal, and then to apply § 172. 

Proof. Suppose that chord QR is II to MAT, thus forming 


Ay in the figure. § 52 

Then x = y, § 61 

and arc PSQ = arc PR. § 155 

Also, y is measured by I arc PR. § 172 

a; is measured by i arc PSQ. Ax. 5 


It may be shown that ZNPQ in the above figure is measured by 
I arc PRQ. This is done by showing that the st. ZNPM is measured 
by half the entire O, and that if we subtract x and h arc QSP, we 
have left ZNPQ and \ arc PRQ. 

It is instructive to consider the arc by which x is measured as 
PQ swings about P, first when PQ is ± to MN and then when PQ lies 
along PN so that a; is a st. Z. 

PS 









136 


FUNDAMENTAL THEOREMS 


BOOK II 


Proposition 17. Two Chords 

178. Theorem. An angle formed by two chords inter- 
secting within a circle is measured by half the sum of 
its intercepted arc and that of its vertical angle. 



Given a O with Z x formed by the chords AC and BD. 

Prove that x is measured by \(arcAB-\- arc CD). 

The plan is to show that x=AC+ZB, and then refer to § 172. 


Proof. Draw BC. Post. 1 

Then x=-Z.C-\-^B. §66 

An exterior Z. of a is equal to the sum of the two 
nonadjacent interior A. 

Also, ZC is measured by J arc AB, 

and ZB is measured by \ arc CD. § 172 

.\x is measured by i (arc AB + arc CD). Ax. 1 


It is interesting to discuss this theorem along the following lines : 

If P is the vertex of Z x, and if we move P to the center of the O, 
to what previous proposition does this one reduce ? 

If P is on the O, as at D, to what previous proposition does this 
one then reduce ? 

Suppose that the point P remains as in the figure, and that the 
chord AC swings about P as a pivot until it coincides with the 
chord BD. What can then be said of the measure of A APB and CPDl 
What can be said as to the measure of A DP A and PPC? 








§§178,179 CHORDS AND SECANTS 137 

Proposition 18. Two Secants 

179. Theorem. An angle formed by two secants^ by a 
secant and a tangenty or by two tangents drawn to a 
circle from an external point is measured by half the 
difference between its intercepted arcs. 



Given two lines PX and PY from an external point P, cutting 
off on a O two arcs b and d such that b'^d. 

Prove that Z.Pis measured by\{b — d). 

The plan is to show that ZP=h'— d', and then to apply §§ 172,177. 

Proof. In the figures as lettered above, we have an 
angle formed by two secants (Case 1), by a secant and a 
tangent (Case 2), and by two tangents (Case 3). 


In each figure draw BD, Post. 1 

In each case, since XlP-{-d' = b', § 66 

we have AP=b'—d'. Ax.2 

Then b' is measured by ^ 6, 

and d’ is measured hy^d, §§ 172,177 

.*. ZP is measured by 1^(6 — d). Ax. 5 


If the secant PY swings around to tangency, it becomes the tan¬ 
gent PB, and Case 1 becomes Case 2. If PX also swings around to 
tangency, it becomes the tangent PD, and Case 2 becomes Case 3. 







138 


FUNDAMENTAL THEOREMS 


BOOK II 


Exercises. Measure of Angles 


1. If two circles are tangent externally and if two line 
segments drawn through the point of 
contact are terminated by the circles, the 
chords which join the ends of these lines 
are parallel. 

This can be proved if it can be shown that /.A 
equals what angle ? To what two angles can these angles be proved 
equal by § 177? Are those angles equal? 

2. If one side of a right triangle is the diameter of a 
circle, the tangent at the point where the a 

circle cuts the hypotenuse bisects the other 
side of the triangle. 

If OM is II to AC, then because BO = OA, what is 
the relation of BM to MC ? The proposition there¬ 
fore reduces to proving that OM is II to what line of 
A ABC? This can be proved if ZBOikf can be shownequal to what angle? 



BMC 




3. The radius of the circle inscribed in an equilateral 
triangle is equal to one third the altitude 
of the triangle. 

To prove this we must show that AR equals 
what line segment ? It looks as if AR might equal 
QR, and QR equal OR. Is there any way of proving 
AORQ equilateral ? of proving AAQR isosceles ? 

4. If two lines are drawn parallel to the sides through 
any point on a diagonal of a square, the points where these 
lines meet the sides lie on the circle whose 
center is the point of intersection of the 
diagonals of the square. 

It can be shown that OY= OZ if what two tri¬ 
angles a,re congruent ? How can you prove these 
triangles congruent? Then how can you prove 
that OF= OX and that OX=OW‘i 


zK 

^c 

(- 

X 

>11 
f N/ 

/ 

- —^ D 











§180 


LOCI 


139 


II. Loci 

180. Meaning of Locus. The Latin word for place is locus, 
the plural of which is loci (usually pronounced 15'si in mathe¬ 
matics). In speaking of the place where certain points lie, 
it is often convenient to speak of it as the locus of the 
points. For example, if a gun at G in this 
figure can be turned through an angle of 
120°, and if the projectile will fall at some 
point between 5000 yd. and 9000 yd., depend¬ 
ing upon the angle at which the gun is ele¬ 
vated, the locus of the points at which the projectile may 
fall is a certain region which is represented by the shaded 
part of the figure. 

While it is proper to represent a locus as a surface, a portion of 
space, a line, or even a point, it is the custom in plane geometry to 
study only those loci which are lines. All the definitions and discus¬ 
sions of loci in Book II refer to loci in a plane. 

The following statements concerning loci are so evident 
that they may be treated as postulates: 

1. The locus of points at a given distance d from a given 

line X is a pair of lines, I and V, I - ir 

parallel to x and at the distance d x _f- 

from it. i'. _i_ 

It is then said that any point on Z or Z' satisfies the con¬ 
dition that it is at the distance d from x. 

Instead of speaking of the locus of points that satisfy a given con¬ 
dition, we may speak of the locus of a point as satisfying the condition. 

2. The locus of points equidistant from a given 
point is a circle whose center is the given point. 

Since the circle is a very obvious locus, the subject 
of loci is considered in Book II. 








140 


LOCI 


BOOK II 


181. Proof of a Locus. To prove that a certain line or group 
of lines is the locus of points that satisfy a given condition, 
it is necessary and sufficient to prove two things: 

1. Every point on the line or lines satisfies the given 
condition; 

2. Every point which satisfies the given condition lies on 
the line or lines. 

If we can prove this for any point whatsoever, that is, not merely 
for some special point, it is evidently true for every point. 

One of the best ways of determining a locus is to take on paper a 
number of points which satisfy the given condition and then try to 
determine on what line or lines they lie. 


Exercises. Loci in a Plane 

State without proof the following loci: 

1. The locus of the tip of the hour hand of a watch. 

2. The locus of the center of the hub of an automobile 
wheel as the car moves straight ahead on a level road. 

3. The locus of the tips of a pair of shears as they open, 
provided the bolt which holds the blades together remains 
always fixed in one position. 

4. The locus of the center of a circle that rolls around 
another circle, inside or outside, and always just touches it. 

Draw the following loci, hut give no proofs: 

5. The locus of points i in. below the base of a given 
A ABC, and also of points i in. above the base. 

6. The locus of points i in. from a given line AB. 

7. The locus of points f in. from a given point O. 

8. The locus of points I in. outside the circle drawn with 
a given point O as center and a radius of 1 in. 


§§ 181,182 


PERPENDICULAR BISECTOR 


141 


Proposition 19. Perpendicular Bisector 

182. Theorem. The locus of points equidistant from two 
points is the perpendicular bisector of the line segment 
joining them. 



Given two points A and and /, the X bisector of AB, 

Prove that every point on I is equidistant from A and B 
and that every point equidistant from A and B lies on I, 


The plan is first to apply § 117. 

Proof. From P, any point on I, draw PA, PB. Post. 1 
Then PA=PB. §117 

This proves the first part of the theorem. 

Let P be any point in the plane such that PA = PB. 

Suppose that PM bisects ZAPB. Post. 8 

Then A AMP is congruent to A BMP. § 40 

.*. AM = BM, and the A at M are equal. § 38 

Hence the A at M are rt. zi, § 13 

and PM is X to AB. § 14 


Since there is only one point of bisection (Post. 7), and 
since only one X can be constructed at M (Post. 10), PM is 
the X bisector oi AB; that is, P lies on I, 

This proves the second part of the theorem. 









142 


LOCI 


BOOK II 


Proposition 20. Bisector of an Angle 

183. Theorem. The locus of points equidistant from two 
given intersecting lines is a pair of lines which bisect 
the omgles formed by them. 



Given two lines AC and BD intersecting at 0, and I and 
the bisectors of AAOB and BOC respectively. 

Prove that every point on I or V is equidistant from AC 
and BD and that every point that is equidistant from 


AC and BD is on I or V, 

The plan is to prove the two statements of § 181. 

Proof. Let P be any point on Z, 
and let PQ be ± to AC and PR be ± to BD. § 116 
Since AAOB is, bisected by Z, Given 

then rt. A OQP is congruent to rt. AORP. § 68 

.*. PQ = PR, or P is equidistant from AC and BD. § 38 
Let P be any point in the plane such that _L PQ = ± PR. 
Draw PO. Post. 1 

Then rt. A OQP is congruent to rt. A ORP. § 71 

ZAOP = ZBOP; §38 

that is, Plies on the bisector of ZAOB, or on Z. Post. 8 


Evidently both parts of the same proof hold for I' and ZBOC. 










§§183-185 INCENTER AND CIRCUMCENTER 


143 


184. Incenter of a Triangle. There are four propositions 
relating to loci which are often given as exercises; but 
because of their special interest they are here given more 
prominence, although their inclusion as fundamental propo¬ 
sitions is optional. The first of these propositions relates 
to the bisectors of the angles of a triangle. 

Theorem. The bisectors of the angles of a triangle 
meet in a point equidistant from the three sides. 

In giving the proof the student should first show that the bisectors 
of A A and B intersect as at O. 
from AC and AB, also from BC 
and ABy and hence from AC and 
BC. Then by § 183 prove that O 
lies on the bisector CZ. 

This point is called the in¬ 
center of the triangle. 

The reason for using this term will appear in § 193, where the prob¬ 
lem which gives this theorem its importance is considered. 

185. Circumcenter of a Triangle. The second proposition 
in this group relates to the perpendicular bisectors of the 
sides of a triangle. 

Theorem. The perpendicular bisectors of the sides of 
a triangle meet in a point equidistant from the vertices. 

In giving the proof, show that the ± bisectors of the two sides BC 
and CA intersect as at O. Then prove 
that O is equidistant from B and C, 
also from A and C, and hence from A 
and B. Then prove that O lies on the 1 
bisector PP'. 

This point is called the circum¬ 
center of the triangle. 

The reason for using this term will appear in the problem of § 188. 



Then prove that O is equidistant 
C 





144 


LOCI 


BOOK II 


186. Orthocenter of a Triangle. The third proposition re¬ 
lates to the altitudes (§ 74) of a triangle. 

Theorem. The altitudes of a triangle meet in a point 

In giving the proof let the 
altitudes be AQ, BR, and CP. 

Through A, B, C draw B'C', 

C'A', and A'B' II to CB, AC, 
and BA respectively. Then 
prove that C'A = BC = AB'. 

What is the relation of AQ to 
B'C' ? In the same way prove 
that BR and CP are the ± bi¬ 
sectors of the other sides of the 
AA'B'C'. Then apply § 185. 

This point is called the orthocenter of the triangle. 

The prefix ''ortho-" means straight, and this center is found by 
drawing lines from the vertices straight (perpendicular) to the sides. 

187. Centroid of a Triangle. The last proposition relates 
to the medians (§ 132) of a triangle. 

Theorem. The medians of a triangle meet in a point 
which is two thirds of the distance from each vertex 
to the midpoint of the oppo¬ 
site side. 

In giving the proof let any two 
medians, as AQ and CP, meet as at 
O. Then if F is the midpoint of AO 
and X of CO, prove that YX and PQ A ^^—^^ 
are II to AC and equal to ^ AC. Then 

prove that A Y=YO= OQ, and that CX = XO = OP. Hence any median 
cuts off any other median two thirds of its length from the vertex. 

This point is called the centroid of the triangle. 

The syllable "-oid " means like, so that the word "centroid" means 
centerlike. This point is the center of gravity of the triangle. 








§§ 186, 187 


EXERCISES 


145 


Exercises. Circular and Straight-Line Loci 


1. The locus of the vertex of a right triangle which has a 
given hypotenuse as hose is the circle constructed upon this 
hypotenuse as diameter. 


2. The locus of the vertex of a triangle which has a given 
base and a given angle at the vertex is the arc which forms 
with the base a segment of a circle in which the given angle 
may be inscribed, 

3. Two forts are placed 28 mi. apart on opposite sides 
of a harbor entrance. Each fort has a gun with a range of 
16 mi. Draw a plan showing the area which can be exposed 
to the fire of both guns, using a scale of in. to a mile. 


4. A straight rod AB moves so that 
its ends constantly touch two fixed rods 
which are perpendicular to each other. 
Find the locus of its midpoint M, 

5. Show how to locate a light equidis¬ 
tant from two intersecting streets and 
48 ft. from the point of intersection, as 
shown in the figure. 



-4- 

/ 


6. A line moves so that it re¬ 
mains parallel to a given line and 
so that one end lies on a given circle. 
Find the locus of the other end. 



7. A circle of center 0 and radius r' rolls around a fixed 
circle of radius r, always touching the fixed circle. What 
is the locus of O ? Prove it. 

8. Find the locus of the mid- 
point of a line segment drawn 
from a given external point to a 
given circle. 












146 


LOCI 


BOOK II 


9. A water main has a gate located at a point 9 ft. from 
a certain lamp-post which stands on the edge of a straight 
sidewalk. The gate is placed 4 ft. from the edge of the 
walk, toward the street. Draw a plan showing every pos¬ 
sible position of the gate and state the principles involved. 

10. During a war a man buried some valuables. He re¬ 
membered that they were buried north of an east-and-west 
line which joined two trees 140 ft. apart, and that the point 
was 80 ft. from the eastern tree and 100 ft. from the western 
tree. Draw a plan to the scale of 20 ft. = 1 in., indicate 
the point where the valuables were buried, and state the 
geometric principles involved. 


11. Find the locus of the center of a circle that passes 
through a given point between two 
parallels and cuts equal chords of a 
given length from them. 

Let P be the given point, AB, CD the 
given parallels, and MN the given length. 

Since the circle cuts equal chords from 
two parallels, what must be the relative distance of its center from 
each ? Then what line is one locus for O, the center of the circle ? 

Construct the perpendicular bisector of MN, cutting XY at S. How 
does SM compare with the radius of the circle ? What is then another 
locus for O? How can we then find O so as to 
satisfy the given conditions? 



R 


' i I 

S:... 


12. Find the locus of a point equidistant 
from two given points P, Q and at a given 
distance d from a third given point R. 

13. Find the locus of the center of a 1 

circle that has a given radius and passes 
through two given points. 

14. What is the locus of the midpoints of a number of 
parallel chords of a circle ? Prove it. 







§§ 188,189 FUNDAMENTAL CONSTRUCTIONS 


147 


III. Fundamental Constructions 
Proposition 21. Circle about a Triangle 

188. Problem. Circumscribe a circle about a given 
triangle. 



Given the A ABC, 

Required to circumscribe a O about /\ABC, 

The plan is to show that the intersection of the ± bisectors of two 
sides of the A is the center of the required O. 

Construction. Construct the Js bisecting the sides AB 
and AC as at M and N respectively. §§ 102,104 

These Js meet, as at 0, or else are II. If ON is II to OM, 
then ON is J_ to AB (§ 63), and hence AB is II to AC (§ 57). 
But this is impossible, since AB and AC form ZA (§ 6). 
With O as center and OA as radius, construct a O. Post. 4 

Then O ABC is the required O. 

Proof. O is equidistant from A and B, 

and O is equidistant from A and C, § 182 

.*. O is equidistant from A, B, C, 

Hence the O O passes through A, B, C, § 134, 6 
189. Corollary. Given a circle or an arc, find the center 
of the circle, 

* Take three points on the © or arc and apply § 188, 








148 


FUNDAMENTAL CONSTRUCTIONS 


BOOK II 


190. Corollary. Through any three given points not lying 
in a straight line one circle and only one can pass. 

The points may be considered as the vertices of a A, and hence a O 
can pass through them (§ 188). 

Since the points are not in a st. line (given), points equidistant from 
A, B, and C in the figure of § 188 must lie on MO and NO {% 182). Since 
these lines can intersect in only one point, O, only one O is possible. 

191. Corollary. Two distinct circles can have at most two 
points in common. 

Because if they have three points in common, they will coincide (§ 190). 

192. Corollary. A straight line can intersect a circle in at 
most two points. 

This corollary, which is essentially § 134,4, is introduced at this point 
because of its analogy to § 191. 

Suppose that the st. line ABC can intersect the O O 
in A, B, C. Then OA = OB = OC{% 134,1). 

Hence ZOCB = Z CBO and Z OB A = ZRAO (§ 42), 
and thus each is less than a rt. Z (§ 65). 

Hence if the supposition were true, we should have 
three equal obliques from O to ABC’, but this is impossible (§ 119). 



Exercises. Constructions 

1. Bisect a given arc. 

2. Upon a given line segment as a chord, construct a seg¬ 
ment of a circle in which a given angle 
may he inscribed. 

Proceed as follows: 

Given the line segment AB and the Z m. 

Required on AB as a chord to construct 
a segment of a Q in which Z.m may he 
inscribed. 

Construction. Construct Z ABX = m (§ 106). 

The rest of the construction is readily inferred from the figure. 





§§ 190-194 


CIRCLES AND TRIANGLES 


149 


Proposition 22. Circle in a Triangle 
193. Problem. Inscribe a circle in a given triangle. 



Given the AA5C. 

Required to inscribe a O in /\ABC. 

The plan is to show that the intersection of the bisectors of two A of 
the A is the center of the required O. 

The construction and proof, which are suggested by the figure, are 
left for the student. 

194. Centers of a Polygon. The center of a circle cir¬ 
cumscribed, if possible, about a polygon is called the 
circumcenter of the polygon. 

The center of a circle inscribed, if 
possible, in a polygon is called the in¬ 
center of the polygon. 

The intersections of the bisectors of 
the exterior angles of a triangle are the 
centers of three circles, each of which is 
tangent to one side of the triangle and to 
the other two sides produced. These three circles are called 
escribed circles, and their centers are called the excenters of 
the triangle. 

f 











150 


FUNDAMENTAL CONSTRUCTIONS 


BOOK II 


Proposition 23. Constructing a Tangent 

195. Problem. Through a given point construct a tan- 
gent to a given circle. 



Given the point P and the O 0, 

Required through P to construct a tangent to the O. 

The plan is to construct a line which shall make a rt. Z. with a radius. 


Construction. 1. If P is on the O, draw OP. Post. 1 

At P construct ' XY ± to OP. § 104 

Then XY is the required tangent. 

2. If P is outside the O, draw OP. Post. 1 

Bisect OP, as at X, § 102 

With X as center and XP as radius, construct a O inter¬ 
secting OO as at Af and AT, and draw PM, Posts. 4, 1 

Then PM is the required tangent. 

Proof. 1. Since XY is ± to OP, Const. 

XY is tangent to the O at P. § 146 

2. Drawing OM, ZPMO is a rt. Z. § 173 

.*. PM is tangent to the O at Af. § 146 


In like manner we may prove that PN is tangent to the ©. 

f 










§195 


EXERCISES 


151 


Exercises. Constructions 



1. If two opposite angles of a quadrilateral are supple¬ 
mentary, the quadrilateral can be inscribed 
in a circle. 

Apply § 188 to constructing a circle through A, B, C. 

Prove that if the circle does not pass through D also, 

ZZ> is greater than or less than some other angle that 
is supplementary to ZR, which is impossible. 

2. In a Is ABC construct PQ II to the base AB and cutting 
the sides in P and Q so that PQ = AP-[-BQ. 

Assume for the moment that the problem is solved. 

Then AP must equal some part of PQ, as PX, 
and BQ must equal QX. 

But if AP = PX, what must ZPXA equal ? 

Since PQ is II to AB, what does ZPXA equal ? 

Then why must Z BAX = Z XAP ? A^^^ — ^ 

Similarly, what about ZQBXand ZXBAl. 

Now reverse the process. What should we do to Z A and B in order 
to fix X ? Then how shall PQ be constructed ? Now give the proof. 



3. Construct a line intersecting two sides of a tri¬ 
angle and parallel to the third side, such that the part 
intercepted between the two sides has a 
given length. 

If PQ = d and if QR is II to PA, what does AR 
equal? Then what two constructions must you ^ 
make in order to locate Q ? B 



4. Construct a triangle, given one side, an adjacent 
angle, and the difference between ^^ c/'' 

the other two sides. a - 

If AB, ZA, and the difference d be- 1 

tween AC and RC are given, what points — A^ - "" ^B 

in this figure are determined ? Can XB be \ 

constructed? What kind of triangle is XBCl How can the vertex C 
of the triangle be located ? 

PS 













152 


FUNDAMENTAL CONSTRUCTIONS 


BOOK II 






5. Given two angles of a triangle and the sum of two 

sides, construct the triangle. c 

Can the third angle be found ? Assume the 
problem solved. If AX=AB-{-BCy what kind 
of triangle is BXC"! What does ZCBA equal? A‘ 

Is ZX known ? How can C be fixed ? 

6. Through a given point P between 

the arms of an zlAOB construct a line 
terminated by the arms of the angle 
and bisected at P. q 

If PM=PNy and PQ is II to PO, is OQ = QikT? 

7. Given the perimeter of a triangle, one angle, and 
the altitude from the vertex of the given angle, construct 
the triangle. 

Assume for the mo¬ 
ment that the problem is 
solved, as shown in this 
figure, in which ABC is 
the required triangle, MN 
the given perimeter, ZACB the given angle, CP the given altitude, 
AM=ACy and BN=BC. By a study of the figure we shall be led to 
the following solution: 

As in Ex. 2, page 148, on MN construct a segment of a circle in which 
ZMCN, which is found by the analysis to be equal to 90°+ \ZACB, 
may be inscribed. Construct XC II to MAT at the distance CP and cut¬ 
ting the arc of the circle at C. Then the vertices A and B are on the 
perpendicular bisectors of CM and CN. 

8. Construct a line that would bisect the angle formed 
by two lines if those lines were pro¬ 
duced to meet. 


X- - 


^’0 

\ 

\ 

\ 

\ 


''''' '' / 

\/ 

\ 

\/ wN. 


A'' 


/p 


■N 



If AB and CD are the given lines, and if 
they could be produced to meet, then the 
bisector of the angle between them would 
be the perpendicular bisector of PQ, a line which makes equal angles 
with the given lines. How can we construct PQ so as to make ZP=ZQ1 









§195 


EXERCISES 


153 


9. Construct a common tangent to two given circles. 



If the centers are O and O' and the radii r and r', the tangent QR 
in the left-hand figure seems to be II to O'M, a tangent from O' to a 
circle whose radius is r — r'. What does this suggest? 

In general, there are four common tangents, but circles tangent 
externally and internally and intersecting circles should be considered. 

Construct an isosceles triangle, given: 

10. The base and the angle at the vertex. 

11. The base and the radius of the circumscribed circle. 

12. The perimeter and the altitude. c 

In this figure ABC is the required triangle 

and MN the given perimeter. Then the alti¬ 
tude CP passes through the midpoint of MN, 
and the A MAC and NBC are isosceles. 

13. Construct an equilateral triangle, 
given the radius of the circumscribed 
circle. 

14. Construct a rectangle, given one side 
and the angle between the diagonals. 

15. Given two perpendicular lines ABand 
CD intersecting in 0, and a line intersecting 
these perpendiculars in E and F, construct a 
square, one of whose angles shall coincide 
with one of the right angles at O, and such 
that the vertex of the opposite angle of the 
square shall lie on EF. 

Notice the two solutions. 











154 


FUNDAMENTAL CONSTRUCTIONS 


BOOK II 


Exercises. Applications 


1. Two pulleys of radii 1 ft. 6 in. and 2 ft. 3 in. respectively 
are connected by a belt which runs straight between the 
points of tangency. If the centers of the pulleys are 6 ft. 
apart, construct the figure, using the scale of 1 in. = 1 ft. 

2. Given a portion of the tire of a 
wheel, show how to determine the center 
and to reproduce the tire of the wheel in 
a drawing. 

3. Construct this design, making the 
figure twice this size. 

First construct the equilateral triangle. Then con¬ 
struct the small cir/^les with half the side of the 
triangle as a radius. Then find the radius of the cir¬ 
cumscribed circle. 

4. A circular window in a church has a 
design similar to the accompanying figure. 

Construct the design, making the figure 
twice this size. 

This design is made from the figure of Ex. 3. 




5. From two given points P and Q construct lines which 
shall meet on a given line AB and make 
equal angles with AB, 

Since Z.BXQ must be equal to Z.AXP, then 
Z.MXP'^Z.MXP. If PP' is ± to AB, so that 
MP' = MP, and if P'Q is drawn, what follows ? 

6. Find the shortest possible path 
from a point P to a line AB and thence 
to a point Q, 

If Z PXA = AQXB, is PX+XQkPR+RQ'I 

This problem shows that a ray of light is re¬ 
flected in the shortest possible path. 



1 


^ M 

i 

If'''" 






§195 


REVIEW EXERCISES 


155 


Exercises. Review 

1. Make a list of the numbered propositions in Book II, 
stating under each the propositions in Books I and II upon 
which it depends either directly or indirectly. 

2. Make another list of the numbered propositions, stat¬ 
ing under each the propositions in Book II 
which depend upon it. 

3. Show how to construct a tangent to this 
circle at the point P, the center of the circle not 
being accessible. 

4. In this figure it is given that 
X = 34° and y — 56°. Find the number 
of degrees in each of the other angles 
and determine whether or not AB is a ^ 
diameter of the circle. 

5. In a circle with center 0 the chord AB is drawn so 
that /-BAO = 31°. How many degrees are there in Z.AOB ? 

6. In this figure it is given that 
ZP=44°, ZA = 76°, and Z.BDC=^h\ 

Find the number of degrees in each of 
the other angles, and determine whether 
or not CD is a diameter. 

7. In a circle with center O the chord AB is drawn so 
that /.BAO = 35°. On either arc AB a point P is taken and 
joined to A and P. What is the- size of /APB ? 

8. Find the locus of the midpoint of a chord formed 
by a secant from a given external point to a given circle. 

9. Show how a carpenter^s square may 
be used to determine whether or not the 
curve in this casting is a perfect semicircle. 

State the geometric principle involved. 











156 


FUNDAMENTAL CONSTRUCTIONS 


BOOK II 


10. In a circle with center 0, OM and ON are con¬ 
structed JL to the chords AB and CD respectively, and it 
is known that Z.NMO = Z ONM. Prove that AB = CD. 

11. Two circles intersect at A and B, and a secant drawn 
through A cuts the circles at C and D. Prove that ADBC 
does not change in size, however the secant is drawn. 

12. Let A and B be two fixed points on a given circle, 
and M and N the ends of any diameter. Find the locus of 
the point of intersection of the lines AM and BN. 

13. Given the sum of the diagonal and one side of a 
square, construct the figure. 

Assuming the problem solved, produce the 
diagonal CA, making AE = AB. Then CE is 
the given sum and ZACB = ZBAC=45°. 

Why ? Find the value of ZE. Reversing the 
reasoning, construct AE and ECB on EC. 

14. If the opposite sides of an inscribed quadrilateral 

are produced to intersect, the bisectors of the angles at 
the points thus found intersect at Q 

right angles. 

Referring only to arcs instead of 
chords, we have 

AX-MD=XB-CM, 
and YA-BN=DY-NC. 

YX+NM=MY+XN. 

Hence ZYIX=ZXIN. 

How does this prove the proposition ? Discuss the impossible case. 

Construct a right triangle^ given: 

15. The median and the altitude upon the hypotenuse. 

16. The hypotenuse and the altitude upon the hypotenuse. 

17. Construct a triangle, given one side, an adjacent 
angle, and the sum of the other sides. 







BOOK III 

PROPORTION AND SIMILARITY 
I. Fundamental Theorems 

196. Proportion. An expression of equality between two 
ratios is called a proportion. 

Preferably, a proportion is written in the more familiar 
fractional form, as follows: 

a_c 
b d 

For convenience in printing, however, the form a:b=c:d, 
or a/b = c/d, is often used. All three forms have the same 
meaning, and each is read ''a is to 6 as c is to d,’" or ''the 
ratio of a to 6 is equal to the ratio of c to 

197. Terms. In a proportion the four quantities compared 
are called the terms. The first and third terms are called 
the antecedents; the second and fourth terms, the conse¬ 
quents. The first and fourth terms are called the extremes ; 
the second and third terms, the means. 

Thus, in the proportion a:b = c:d, a and c are the antecedents, 
6 and d the consequents, a and d the extremes, h and c the means. 

Such names were of more value before algebra came into common 
use than they are at present. 

In the case of a:b = b:c, the term 6 is called the mean 
proportional between a and c. 

There is only one positive mean proportional between two numbers, 
and hence we speak of the mean proportional, as above, 

167 


158 


FUNDAMENTAL THEOREMS 


BOOK III 


198. Algebraic Relations. Since we are treating of the 
numerical measures of lines, we shall treat all ratios alge¬ 
braically. The following laws should be understood : 

1 . If ^ = ^ then ad = he, 

0 a 

For we may multiply each of the given equals by hd. Ax. 3 


2. If ^ = then d = b; and if Y = then a = c. 

0 d 0 0 

For ad = ab, by the first law, and hence d = b; or ab = cb, and 


hence a = c. 


S. If ad = bCy then 7 = v 
0 d 

For we may divide each of the given equals by bd. 


Ax. 4 


Ax. 4 


4 . If f = then^ = ^- 

b d c d 

For we may multiply each of these equals by ^ • Ax. 3 

5 . If 2 = £, then ^ = 

b d a c 

For we may divide 1=1, member for member, by these equals. Ax. 4 

6 . = = 

b d b d 

For we may add 1 to each of these equals, giving ^ ^ ^ = UlA . Ax. 1 

0 d 

7 . Jf | = £, then = 
b d b 

For we may subtract 1 from each of these equals. 


d 


a + c+e4-j7+- 


Ax. 2 
a 


8 If —d. - r then ■ ■ ■— r 

For a = br, c = dr, e=fr, g = hr,’^‘, and hence 

a + c + e + 5r + ... = r(6 + d +/+ /H- • • •); 


, a + c + e + gr + -’* a 

whence , , , . ^ ^ . -= r = -• 

b + d-\-f + h‘ b 


Ax.] 
Ax. 4 








§198 


LAWS OF PROPORTION 


159 


Exercises. Algebraic Relations 

Prove the following as in § 198 or by referring to §198: 

1. In any proportion the product of the extremes is 
equal to the product of the means. 

2. If the two antecedents of a proportion are equal, 
the two consequents are equal. 

3. If the product of two quantities is equal to the product 
of two others, either two may be made the extremes of a 
proportion in which the other two are made the means. 

4. If four quantities are in proportion, they are in pro¬ 
portion by alternation ; that is, the first term is to the third 
as the second term is to the fourth. 

5. If four quantities are in proportion, they are in pro¬ 
portion by inversion ; that is, the second term is to the first 
as the fourth term is to the third. 

6. If four quantities are in proportion, they are in pro¬ 
portion by composition'; that is, the sum of the first two 
terms is to the second term as the sum of the last two 
terms is to the fourth term. 

7. If four quantities are in proportion, they are in pro¬ 
portion by division; that is, the difference between the 
first two terms is to the second term as the difference 
between the last two terms is to the fourth term. 

8. In a series of equal ratios the sum of the antecedents 
is to the sum of the consequents as any antecedent is to 
its consequent. 

9. li a;h = c:dj then a^:b^=c^: d^, 

10. If a:b = b:c, then a:c = a^ 

11. \i a\b = b:Cy then b = ^ac. 

We shall consider only positive numbers unless the contrary is stated. 


160 


FUNDAMENTAL THEOREMS 


BOOK III 


199. Nature of the Quantities in a Proportion. Although 
we may have ratios of lines, of areas, or of other geometric 
magnitudes, we treat all the terms of a proportion as 
positive numbers. 

If b and d are lines or solids, for example, we cannot 
multiply each member of | ^ by bd, as in § 198,1, because 

we cannot think of multiplying by a solid. 

Hence when we speak of the product of two geometric 
magnitudes, we mean the product of the numbers which 
represent the magnitudes when they are expressed in terms 
of a common unit, 

200. Proportional Line Segments. If we have two line seg¬ 
ments AB and A'B' and if M and M' are their respective 
midpoints, then AM\MB = 1, and 

A'M': M'B' = 1, and hence A --- B 

AM _ A'M\ _ Ml _ B' 

MB M'B' 

This is evidently true whatever may be the lengths of 
AB and A'B'. 

In like manner, if we have two line segments XY and 
X'Y', we may divide XY at P and X'Y' at P' in such a 


way that 



Y 


XP _ X'P' 
PY P'Y' 


When we divide two line segments in such a way as to 
have the parts form a proportion like this one, we say that 
the line segments are divided proportionally. 

If P is on the line AB and is between A and B, it divides 
AB internally ; if it is not between , 

A and B, it divides AB externally. * --- S. 

In this figure, P divides AB internally in the ratio 1:2, and P' divides 
AB externally in the ratio 1:2. That is, AP : PB = AP ': P'B = 1:2, 







§§ 199,200 


PROPORTION 


161 


Exercises. Proportion 

Express the following ratios in their simplest forms: 

1. 10:12. 4. a:al 7. f:|. 10. a:a^ + a6. 

2. 8a:12a. 5. 6:9ml 8. |:|. 11. a^-{-ah:a, 

2 32 X ^ g a -\-h ^ q — If ^ ^^2 a^ ~}~ 2 a -}-1 ^ 

* 48x ' a^—h^ ' a —b ’ a +1 


Given the proportion a : h 

13. a:d = bc: d^, 

14. l:b = c:ad. 

15. ad:b = c:l. 

16. ma : b = me: d. 


= c:d, prove the following: 

17. ma:nb = mc:nd. 

18. a—l:b = bc — d:bd. 

19. a+l :l = bc-\-d:d. 

20. l:bc=l:ad. 


21. Divide a line segment 4.2 in. long into two parts 
which shall have the ratio 1:2. 

22. Divide a line segment 3.6 in. long into two parts such 
that the ratio of the shorter part to the whole segment 
shall be 4:5. 

23. What is the ratio of half a right angle to one eighth 
of a straight angle ? 

Given the proportion a:b = b:c, prove the following: 

24. c:b = b:a. 26. (6 + \/^)(6-V^) = 0. 

25. a:c = b^:c\ 27. ac-l:6-l = 6 + l:l. 

Find the value of x in each of the following: 

28. 2:8 = a::12. 30. 7:a; = x:28. 

29. 3:5 = ir:9. 31. l:l+a; = a:-l:3. 

Certain exercises on this page, such as Exs. 1-20 and 24-31, are 
introduced merely for the purpose of accustoming the student to the 
use of ratios and proportions. They are not needed in geometry, and 
may therefore be omitted if desired. 





162 


FUNDAMENTAL THEOREMS 


BOOK III 


Proposition 1. Sides of a Triangle 

201. Theorem. If through two sides of a triangle a 
line is constructed parallel to the third side, it divides 
the two sides proportionally. 



Given the A ABC with PQ II to AB. 


Prove that 


AP^BQ, 
PC QC 


Proof. Assuming AP and PC commensurable, let some seg¬ 
ment AM be contained 3 times in APand 4 times in PC. § 163 


Then 


AP_ 3 ♦ AM _3 
PC 4 * AM 4‘ 


§165 


At the several points of division on APand PC construct 
lines II to AB. § 107 

These lines divide PC into 7 equal parts, of which BQ 
contains 3 parts and QC contains 4 parts. § 85 


Then, 

Hence 


BQ_S 
QC 4* 
AP _BQ 
PC QC 


§165 
Ax. 5 


The proof is evidently the same if any other numbers are used. 

For the incommensurable case (§ 166) see § 516. 

Since the student is now so far advanced as to be able to state for 
himself the plan of attack, it is no longer given as part of the printed 
proof. The student, however, should give it as part of his proof. 


PS 














§§ 201-203 


SIDES OF A TRIANGLE 


163 


202. Corollary. One side of a triangle is to either of its 
segments cut off by a line parallel to the base as the third 
side is to its corresponding segment. 


In the figure of § 201, 


AP BQ 
PC~ QC 

Adding 1 to each member of this proportion, we have 

^+1 = ^ + 1 
PC QC ’ 


§201 


Ax. 1 


or 

AP+PC BQ+QC 


PC QC * 


whence 

AC BC 

PC QC 

Ax. 10 

We may also begin with 

AP BQ 

§ 198, 6 

AC BC 

add 1 as above, and end with —— =- 



AP BQ 



203. Corollary. Three or more parallel lines cut off pro¬ 
portional segments on any two transversals. 


Construct 

AN 11 to CD. 

§107 

Then 

AL = CC, 

LM= GK, 


and 

MN= KD. 

§78 

Now 

AF_AL . 
FH~ LM' 

§201 

Hence 

II 

Ax. 5 

or 

gfe 

II 

C^ie? 

§ 198, 4 




That is, the first two segments of AB are proportional to the first 
two segments of CD. Similarly, the other segments are proportional. 
This is indicated as follows : 

AF _FH ^HB ^ 

CG~GK KD ' 


The student should also consider the case in which AB and CD 
intersect between AC and BD. 








164 


FUNDAMENTAL THEOREMS 


BOOK III 


Proposition 2. Converse of § 201 

204. Theorem. If a line divides two sides of a tri¬ 
angle 'proportionally from a vertex, it is parallel to 
the third side. 



Proof. Taking the indirect method (§ 56), suppose that 
PQ is not II to AB. 

From P construct some other line, as PXy II to AB. § 107 


Then 

^_BC 

PC xc' 

§202 

But 

AP_BQ 

PC QC' 

Given 

Hence 

AP + PC BQA-QC 
^ ^ — ■■ .» 

§ 198, 6 


PC QC 

or 

AC BC 

Ax. 10 


PC QC 

Hence 

BC BC 

Ax. 5 


QC XC' 

and 

QC = XC. 

§ 198,2 


/. PQ and PX must coincide. 

Post. 1 

But 

PX is II to AB. 

Const. 


, PQ, which coincides with PX, is II to AB, 

§52 












§204 


SIDES OF A TRIANGLE 


165 


Exercises. Proportional Lines 


1. In the figure of §203 given that AF= 2 in., FH= 3 in., 
and CK = 6 in. Find the length of CG, 

2. If a side of this square is 10 in., the 
diagonal DB is 14.14 in. long. If DP = 4 in. 
and PQ is II to ABy what is the length of DQ ? 

3. The sides of a triangle are 6 in., 8 in., 
and 10 in. respectively. A line parallel to the 8-inch side cuts 
the 6-inch side 2 in. from the vertex of the largest angle. 
Find the lengths of the segments of the 10-inch side. 



4. Two joists 6 in. wide are fitted together at right angles, 
as here shown. The distance from A to ^ is 16 ft., that from 
A to C is 12 ft., and that from to C is 
20 ft. In fitting another joist along the 
dotted line BC the carpenter has to saw 
off the- ends of the first joists on the 
slant. Find the length of the slanting 
cut across the upright piece; across the horizontal piece. 




5. From any point P the lines PA, PP, PC are drawn to 
the vertices of a AAPC and are bisected respectively by 
A', B\ and C'. Prove that ZCPA = ZC'P'A'. 

6. From any point P within the quadrilateral ABCD lines 
are drawn to the vertices A, P, C, D and are bisected by 
A', P', C'y D\ Prove that ZCPA = ZC'B'A’. 

7. If a spider, in making its web, makes A'B' II to AP, 
P'C' II to PC, C'P' II to CP, D'E' II to DEy 
and E'F' II to EFy and then runs a line 
from F' II to PA, will it strike the 
point A'? Prove it. 

First show that OA': AA = OF': F'F, and 
then use § 204. 










166 


FUNDAMENTAL THEOREMS 


BOOK III 


205. Similar Polygons. Polygons that have their cor¬ 
responding angles equal and their corresponding sides 
proportional are called 
similar polygons, / 

Thus, the polygons ARCZ)-E7 j 

and A'B'C'D'E' are similar if ^ - 'b 

ZA = ZA', ZB = ZB', Z.C = AC\ ZD = ZD\ ZE = ZE\ 

AB ^ BC ^ CD _ DE ^ EA 
A'B'~B'C' CD' D'E'~ E'A'' 



Instead of saying that two polygons are similar, it is frequently 
said that they have the same shape, or that they are the same figure 
drawn to different scales. Familiar illustrations of similar polygons 
are given by maps or by photographs of buildings. 


Q 


Q' 


In the figures here shown, Q and Q' are not similar, 
for although the corresponding sides are proportional, the 
corresponding angles are not 
equal; neither are the figures 
R and R' similar, even though 
the corresponding angles are 
equal, for the corresponding 
sides are not proportional. 

As will be shown in §§208-214, in the case of triangles 
either condition implies the other, but this is not true of 
other figures. 


[ZJ mj 


206. Corresponding Line Segments. In similar polygons 
those line segments that are similarly situated with respect 
to the equal angles are called corresponding line segments^ 
or simply corresponding lines. 

Corresponding lines are occasionally called homologous lines. 


207. Ratio of Similitude. The ratio of any two corre¬ 
sponding line segments in similar polygons is called the 
ratio of similitude of the polygons. 











§§ 205-207 


SIMILAR POLYGONS 


167 


Exercises. Review 

1. If a pendulum swinging from the point O cuts two 
parallel lines at the varying points P and Q respectively, 
the ratio OP : OQ remains the same whatever may be the 
position of the pendulum. 

2. Through a fixed point P a line is drawn cutting a 
fixed line at X. The line segment PX is then divided at Y 
so that the ratio PY : YX always remains the same. Find 
the locus of the point F as X moves along the fixed line. 

3. Given that 3: a; = a;: 27, find the value of x, 

4. Given that a;: 8 = 32: a;, find the value of x, 

5. From the definition of a square, prove that two 
squares are always similar. 

6. From what you have proved concerning equilateral 
triangles, can you state that two equilateral triangles are 
always similar ? Give the reasons. 

7. Divide a line segment 5.4 in. long into two parts 
which shall have the ratio of 4 to 5; of 8 to 10; of 2 to 2l. 

8. The law of levers states that mW = nP, where W, as 

in this figure, is the weight; m, the dis- xr 

tance from the weight to the fulcrum F; 

P, the power applied; and ti, the distance P 

from the power to the fulcrum. State 
this equation in the form of a proportion. 

9. From the point P on the side CA of the A ABC par¬ 
allels are drawn to the other sides, meeting AB in Q and BC 
in R. Prove that A Q: QB = BR : RC. 

10. In the AABC the points P and Q are taken on the 
sides CA and BC so that AP:PC = BQ:QC. Then a line 
AR is drawn II to PP, meeting CB produced in P. Prove 
that CQ:CB = CB:CR, 

PS 



168 


FUNDAMENTAL THEOREMS 


BOOK III 


Proposition 3. Mutually Equiangular Triangles 

208. Theorem. Two mutually equiangular triangles 
are similar. 



Given the AABC and A'B'C with AAy By C equal to 
AA\B'y C’ respectively. 

Prove that A ABC is similar to A A'B'C'. 


Proof. Since the A are given as mutually equiangular, 
we have only to prove that 


AB _BC _ AC _ 
A'B' B'C' A'C'' 


§205 


Place AA'B'C' upon AABC so that AC' coincides with 
its equal, AC, and A'B' takes the position PQ. Post. 5 
Then, in the figure, p = Z A, Given 

and hence PQ is II to AB. § 59 


Then 


AC 

PC 


.f(S2te).or 


AC 

A'C' 


BC 

B'C'' 


Ax. 5 


Similarly, by placing AA'B'C' upon AABC so that AB' 
coincides with its equal, AB, we can prove that 


AB _ BC . 
A'B' B'C'" 


whence 


AB _ BC _ AC 
A'B' B'C' A'C' 

AABC is similar to AA'B'C'. 


Ax. 5 
§205 











§§ 208-212 


SIMILAR TRIANGLES 


169 


209. Corollary. Jf two angles of one triangle are equal 
respectively to two angles of another, the triangles are 
similar. 

Since in each A the sum of the A is 2 rt. A (§ 65), and since two A 
of one A are given equal to two A of the other, the third A are equal 
(Ax. 2); that is, the A are mutually equiangular. Hence the A are 
similar (§ 208). 

210. Corollary. If an acute angle of one right triangle is 
equal to an acute angle of another, the triangles are similar. 

Since the rt. A are also equal (Post. 6), the A have two A of one equal 
respectively to two A of the other. Hence the A are similar (§ 209). 

211. Corollary. If two triangles have their sides respec¬ 
tively parallel to one another, the triangles are similar. 

In this figure how can it be proved that 
AB — AB' and that AA = AA'l. Is this sufficient 
to prove the corollary ? 

Although §§211 and 212 are interesting corol¬ 
laries of § 208, they are not needed in subsequent ^ 
propositions. Hence they may be treated as 
exercises or omitted if desired. These corollaries are required in some 
courses of study and are often given in examinations. 

212. Corollary. If two triangles have their sides respec¬ 
tively perpendicular to one another, 
the triangles are similar. 

In this figure, what Z is the complement 
of Z 5 ? of AB' ? Are these two complements 
equal? Does this prove that AB'=^AB‘l ^ 

Since A'B' is _L to AB and A'C' is ± to AC, 
what can you say about the other two A of 
the quadrilateral formed by these lines? 

What other Z is a supplement of one of these A ? Can it then be proved 
that Z A'= Z A ? Is this sufficient to prove the corollary ? 

When, as in this case, a figure becomes somewhat complicated, it is 
well to recall this fact: The corresponding sides of similar triangles are 
opposite the corresponding and mutually equal angles, and conversely. 








170 


FUNDAMENTAL THEOREMS 


BOOK III 


Proposition 4. Angle and Proportional Sides 

213. Theorem. If two triangles have an angle of 
one equal to an angle of the other and the including 
sides proportional, the triangles are similar. 



CA CB 
C'A'~ CB'* 

Prove that AABC is similar to AA'B'C\ 

Proof. Place AA'R'C'upon AABC so that Z.C' coincides 


with its equal, ZC, A'b! taking the position PQ. 

Post. 5 

Now 

CA CB . 

Given 


C'A' C'B'" 


that is, 

CA_CB, 

CP cq' 

Ax. 5 

Hence 

CA-CP CB-CQ^ 

CP CQ 

§ 198, 7 

or 

II 

0|0 

olta 

Ax. 5 


PQ is II toAB. 

§204 

Then 

ZA = p, and Z.B = q. 

§62 

Also, 

ZC = ZC'. 

Given 

Hence 

AABC is similar to APQC; 

§208 

that is. 

A ARC is similar to AA'B'C\ 

Ax. 5 











§§ 213, 214 


SIMILAR TRIANGLES 


171 


Proposition 5. Proportional Sides 

214. Theorem. If two triangles have their sides respec¬ 
tively proportional^ they are similar. 



Prove that AABC is similar to AA'B'C', 

Proof. On CA take CP= C'A' and on CB take CQ = C'B\ 
Draw PQ. Post. 1 

When it is desired to give a considerable number of steps on a 
single page, the fraction form of the proportion may be replaced by 
the form used below. 


Now 

CA:C'A' = BC:B'C’, 

Given 

and, since 

CP = C'A\ and CQ = C'B\ 

Const. 

then 

CA:CP = CB:CQ. 

Ax. 5 


AABC and PQC are similar. 

§213 

Then 

CA:CP = AB:PQ; 

§205 

that is, 

CA:C'A’ = AB:PQ. 

Ax. 5 

But 

CA:C'A' = AB:A'B\ 

Given 


AB:PQ = AB:A’B\ 

Ax. 5 

and 

PQ = A’B', 

§198,2 

Hence 

APQC and A'B'C' are congruent. 

§47 

But 

AABC is similar to APQC, 

Proved 


AAiSCis similar to AA'B'C\ 

Ax. 5 










172 


FUNDAMENTAL THEOREMS 


BOOK III 


Proposition 6. Right Triangle 

215. Theorem. The perpendicular from the vertex of 
the right angle of a right triangle to the hypotenuse 
divides the triangle into two triangles which are similar 
to the given triangle and to each other. 



Given the rt. A ARC with CP _L to the hypotenuse AB. 
Prove that AABC^ ACPj CBP are similar. 

Proof. Lettering the figure as shown, since a is common 
to rt. A ACP and ABC, these A are similar. § 210 

Likewise, ACBP is similar to A ABC. § 210 

Hence A ACP and CBP are each mutually equiangular 
with the given A ABC. § 205 

Then the three A are mutually equiangular, Ax. 5 
and hence the A are similar. § 208 

216. Corollary. The perpendicular from the vertex of the 
right angle to the hypotenuse of a right triangle is the mean 
proportional between the segments of the hypotenuse. 

Since A ACP and CRP are similar, §215 

we have AP : CP = CP: PB, § 205 

and hence the ± CP is the mean proportional between the segments 
AP and PB. § 197 









215-219 


RIGHT TRIANGLE 


173 



217. Corollary. The perpendicular from any point on a 
circle to a diameter of the circle is the q 
mean proportional between the segments 
of the diameter. 

Since ZACBis a rt. Z (§ 173), A ARC is a rt. A, A 
and hence § 216 applies. 

218. Corollary. The square of the hypotenuse of a right 
triangle is equal to the sum of the squares of the other 
two sides. 

This means that the square of the numerical measure of the hypot¬ 
enuse is equal to the sum of the squares of the numerical measures 
of the two sides. 

This is the most celebrated single proposition in geometry, and on 
account of its great importance we shall prove it again, by another 
method, in § 252. This theorem was known for special cases as early 
as the third millennium B.C., but it is thought to have been first proved 
by Pythagoras, a famous Greek mathematician, about 525 B.c. 

In the rt. AARC, in which ZC is the rt. Z, let the X p from C 
to AB form the segments x and y as here shown. Then a simple 
proof, based on § 215, is as follows: 

Since the three A are similar. 


we have 

Hence 

and 

Then 


c a j c 6 
- = -» and - = - 
ay ox 

2 _ 


215 

205 


a" = cy, 

= cx. § 198,1 

4- 52 = c (aj + y). 

. *. + 6^ = c. c = c^. 



219. Projection. If from the ends of a given line segment 
perpendiculars are constructed to a 
given line, the segment thus formed 
on the given line is called the pro¬ 
jection of the given segment upon 
the line. 

Thus A'R' and AR' in these figures are the projections of AR upon 
the lines m and n respectively. 











174 


FUNDAMENTAL THEOREMS 


BOOK III 


Exercises. Similar Triangles 


1. If a perpendicular is drawn from the vertex of the 
right angle of a right triangle to the hypotenuse^ each of the 
other sides is the mean proportional between the hypotenuse 
and the projection of that side upon it. 


2. The squares of the two sides of a right triangle are pro¬ 
portional to the projections of the sides upon the hypotenuse. 


In the figure of § 215, AC = 


Hence 


AR-AP, and^" 
AB^AP AP 
AB-BP BP' 


= ABBP. 


Why? 


3 . The square of the hypotenuse and the square of either 
side of a right triangle are proportional to the hypotenuse 
and the projection of that side upon it. 

In the figure of § 215, AP^= AP- AP, and AC^=AB>AP. 

AB^ _AB>AB _AB 
AC^~ ABAP~ AP' 


Then 


4. If a perpendicular is drawn from any point on a circle 
to a diameter, the chord from that point to either end of 
the diameter is the mean proportional between the diameter 
and its segment adjacent to the chord, 

5. Perpendiculars drawn from any corresponding vertices 
of two similar triangles to the opposite sides have the same 
ratio as any two corresponding sides, 

6 . Find the length of the hypotenuse of a right triangle 
of which the two sides including the right angle are 37 in. 
and 49i in. respectively. 

7. Find the other side of a right triangle of which the 
hypotenuse is 17 in. and one side is 10.2 in. 

8 . From the three similar triangles in the figure of § 215 
it is possible to write a large number of proportions. Write 
twelve of them. 




5§ 220,221 INTERSECTING CHORDS 175 

Proposition 7. Intersecting Chords 

220. Theorem. If two chords of a circle intersect, the 
product of the segments of either one is equal to the 
product of the segments of the other. 



Given a O with the chords AB and CZ>, intersecting at P. 
Prove that PA • PB = PC • PD. 

Proof. Draw AC and DD. Post. 1 

Then in the figure, as lettered above, 

a = a\ § 172 

because each of these A is measured by \ arc CB ; 

and h = h\ § 172 

because each of these A is measured by \ arc DA. 

.*. ACPA and BPD are similar, § 209 

and hence % = 

PA-PB = PC-PD. §198,1 

221. Secant to a Circle. When we speak of a secant from 
an external point to a circle it is understood that we mean 
the segment of the secant which lies between the given 
external point and the second, or more remote, point of 
intersection of the secant and the circle. 

Thus in the figure of § 222 we may speak of BA as such a secant. 








176 


FUNDAMENTAL THEOREMS 


BOOK III 


Proposition 8. Secant and Tangent 

222. Theorem. If from a point outside a circle a secant 
and a tangent are drawn, the tangent is the mean pro¬ 
portional between the secant and its external segment 



Given a secant PA and the tangent PB drawn to the Q)ABC 
from the external point P. 


Prove that 

§15 

II 


Proof. Draw 

AB and BC. 

Post. 1 

Now 

a is measured by \ arc BC, 

§172 

id 

a' is measured by | arc BC. 

§177 


.'. a = a\ 

Ax. 5 

Then 

A PAP is similar to APBC. 

§209 

Hence 

PA_PB 

PB PC 

§205 


223. Corollary. If from a point outside a circle two or 
more secants are drawn, the product of any secant and its 
external segment is equal to the product of any other secant 
and its external segment. 

Since PA :PB = PB: PC (§ 222), then PA PC = FE^{% 198,1). 

Moreover, since PB always remains the same (§ 149), PA-PC always 
remains the same. 









§§222-224 


SIMILAR POLYGONS 


177 


Proposition 9. Ratio of Perimeters 

224. Theorem. The perimeters of two similar polygons 
have the same ratio as any two corresponding sides. 



Given the two similar polygons ABODE and A^B'CD'E^ with 
perimeters p and respectively. 


Prove that 


V _ AB _ 
p~ A'B'' 


Proof. 

Then 


AB _ BC _ CD _ DE _ EA 
A'B' B'C' CD’ D’E' E'A'^ 
AB-\-BC-\-CD-\-DE-\-EA _ AB 

A'B'+ B'C-\-CD'-{- D 'E’-h E'A' A’B' 


§205 
§ 198,8 


Hence Ax. 5 

p AB 

The proof is evidently the same whatever the number of sides of 
the polygons. 


Exercises. Review 


1. If two chords intersect within a circle, their segments 
are reciprocally proportional. 

This means, for example, that, in the figure of § 220, PA : PD is equal 
to the reciprocal of PB : PC ; that is, it is equal to PC : PB. 

2. Discuss § 220 when P is on the circle; when P is out¬ 
side the circle. 










178 


FUNDAMENTAL THEOREMS 


BOOK III 


3. If two parallels are cut by three transversals which 
meet in a point, the corresponding segments of the parallels 
are proportional. 

4. The base and altitude of a triangle are 30 in. and 14 in. 
respectively. If the corresponding base of a similar triangle 
is 7^ in., find the corresponding altitude. 

5. The point P is any point on the arm OX of the Z.XOY, 
and from P a _L PQ is constructed to OY. Prove that for 
any position of Pon OX the ratio OP: PQ remains the same, 
and the ratio PQ : OQ also remains the same. 

6. In drawing a map of a triangular field with sides of 
75 rd., 60 rd., and 50 rd. respectively, the longest side is 
made 2 in. long. How long are the other two sides made ? 


7 . This figure represents part of a diagonal scale some¬ 
times used by draftsmen. Between vertical lines 1 and 0 
the distance is 1 in.; be- 97531 
tween vertical line 1 
and the intersection of 
diagonal line 2 with 
horizontal line 0 it is 
1.2 in.; between verti¬ 
cal line 1 and the inter¬ 
section of diagonal 


% 


line 0 with horizontal line 8 it is 1.08 in.; and so on. Show 
how to measure 1.5 in.; 1.25 in.; 1.03 in.; 1.67in.; 1.79in. 
Upon what proposition does this depend? 

8. In the similar A ABC and A'B'C\ AP=3in., 
BC = si in., CA = 4 in., and A'B'= in. Find P'C' and C'A'. 

9. The perimeter of an equilateral triangle is 72 in. 
Find the side of an equilateral triangle of half the altitude. 

10. The hypotenuse of a right triangle is 98.5 mm. and 
one side is 78.8 mm. Find the other side. 















§225 


SIMILAR POLYGONS 


179 


Proposition 10. Separating Similar Polygons 

225. Theorem. If two 'polygons are similar, they can 
he separated into the same number of triangles, similar 
each to each and similarly placed. 



Given two similar polygons ABCDE and A'B'CD'E^. 

Prove that ABCDE and A'B'C'D'E' can be separated into 
the same number of A, similar each to each and similarly 
placed. 


Proof. Draw DA, D 'A ' and DB, D 'B Post. 1 
thus separating each polygon into three A similarly placed. 
Since /.E = /LE\ 


and 

DE:D'E'=EA:E'A', 

§205 

we see that 

ADEA and D'E'A' are similar. 

§213 

In like manner, ADBC and D'B'C' are similar. 
Furthermore, Z A = Z A', 


and 

^DAE = ZD'A'E'. 

§205 

Subtracting, 

ZBAD = ZB'A'D'. 

Ax. 2 

Now 

DA-.D'A'=EA:E'A', 


and 

AB:A'B'=EA:E'A'. 

§205 

Then 

DA:D'A' = AB:A'B’. 

Ax. 5 

• • 

ADAB and D'A'B' are similar. 

§213 









180 


FUNDAMENTAL THEOREMS 


BOOK III 


Exercises. Review 


1. The sides of a polygon are 3 in., 3 in., 4 in., 4 in., and 
6 in. respectively. Find the perimeter of a similar polygon 
whose longest side is 9 in. 


2. In drawing a map to the scale of 1:100,000, what 
lengths, to the nearest 0.01 in., should be taken for the sides 
of a rectangular county 30 mi. long and 20 mi. wide ? 


3. By adjusting the screw at O, the lengths OA and 

OC of these 'proportional co'mpasses, ^ * _ ,d 

and the corresponding lengths OB 
and OD, may be varied proportionally. ^ ^ 

The distance AB is what part of CD 

when OA = 4| in. and OC = in. ? when OA = 5.25 in. and 

OC = 6.75 in. ? 



4. A baseball diamond is a square 90 ft. on a side. What 
is the distance, to the nearest 0.1 ft., from first base to 
third base ? 


5. Find a formula for the height of an equilateral triangle 
of perimeter p. 


6. Find the lengths of the sides of an isosceles triangle 
of perimeter 39 in. if the ratio of one of the equal sides to 
the base is f. 

7. Find the length and the height of a rectangle of 
perimeter 64 in. if the ratio of these dimensions is f. 


8. Within the polygon ABCDE here shown any point P is 
joined to the vertices. Beginning at 
a point A' on AP lines are drawn so 
that A'B’ is II to AB, B'C' is II to BC, 

C'D' is II to CD, and D'E' is II to DE. 

Prove that a line E'A' is II to EA and 
that the two polygons are similar. 



§226 


SIMILAR POLYGONS 


181 


Proposition 11. Condition of Similarity 

226. Theorem. If two polygons are composed of the 
same number of triangles^ similar each to each and 
similarly placed, the polygons are similar. 



Given two polygons ARCDJS^and A'B^CD'E' composed of the 
ADEA,DAB, Z)5C similar respectively to the AD'E'A\D'A'B', 
D'B'C, and similarly placed. 


Prove that ABODE is similar to A'B'C'D'E', 
Proof. Since ZDAE=ZD'A'E\ 

and since Z BAD = ZB'A'D\ 

because the A are given similar^ 


we have 
Similarly, 
and 
Also, 
and 


ZBAE=ZB'A'E’. 
ZCBA = ZC'B'A\ 
ZEDC=ZE'D’C', 
ZC = ZC\ 
ZE=ZE\ 


§205 

Ax.l 


§205 


Hence the polygons are mutually equiangular. 


Also, 


DE 

D’E’ 


Hence 


EA _pA ^ AB _ DB _ BC _CD §205 
E’A’ D’A' A'B' D'B' B’C' C'D'' 

because the A are given similar. 

the polygons are similar. § 205 









182 


FUNDAMENTAL THEOREMS 


BOOK III 


Proposition 12. Bisector of an Interior Angle 

227. Theorem. The bisector of an angle of a triangle 
divides the opposite side into segments which are pro¬ 
portional to the adjacent sides. 



Given the bisector of ZC of the AABCy meeting AB at M, 



Proof. 

From A construct a line II to MC. 

§107 

Then 

this line must meet BC produced, 
because CM and CB cannot both be W to it. 

§52 

Let 

this line meet BC produced at X. 


Then 

AM:MB=XC:a. 

§201 

Also, 

ZACM=ZCAX, 

§61 

and 

ZMCB = ZX. 

§62 

But 

ZACM=ZMCB. 

§11 


zcAx=zx, 

Ax. 5 

and hence XC=b. 

§69 


Substituting h for XC in the above proportion, 


AM b 








§§ 227, 228 


BISECTORS OF ANGLES 


183 


Proposition 13. Bisector of an Exterior Angle 

228. Theorem. If the bisector of an exterior angle of a 
triangle meets the opposite side produced^ it divides that 
side externally into segments which are proportional 
to the adjacent sides. 



Given the bisector of the exterior Z.XCA of the A ABC, 
meeting BA produced at M'. 


Prove that 


AM' _b 
M'B a 


Proof. 

Construct APW to M'C, meeting BC at P. 

§107 

Then 

M'B:AM'=a:PC, 

§202 

or 

AM':M'B = PC: a. 

§ 198, 5 

Now 

ZXCM'=ZCPA, 

§ 62 

and 

ZM'CA^ZPAC. 

§61 

But 

ZXCM' = ZM'CA. 

§11 


ZCPA=ZPAC, 

Ax. 5 

and hence b = PC. 

§69 

Substituting b for PC in the second proportion, 



AM' b. 

Ax. 5 


M'B a 

What follows when CA = CB ? when CA > CB-? 


FS 









184 


FUNDAMENTAL THEOREMS 


BOOK III 


Exercises. Review 

1. If two circles are tangent externally, the corre¬ 
sponding segments of two lines drawn through the point 
of contact and terminated by the circles are proportional. 

2. If two circles are tangent externally, their common 
external tangent is the mean proportional between their 
diameters. 

3. Two circles are tangent, either internally or exter¬ 
nally, at P. Through P three lines are drawn, meeting one 
circle in X, Y, Z and the other in X\ Y\ Z' respectively. 
Prove that ^XYZ and X'Y'Z' are similar. 

4. If two circles are tangent internally, all chords of the 
greater circle drawn from the point of contact are divided 
proportionally by the smaller circle. 

5. From a point P a secant 7.6 in. long is drawn to a 
circle such that the external segment is 1.9 in. Find the 
length of the tangent from P. 

6. In a A ABC, AB = 16, BC = 14, and CA = 15. Find the 
segments of CA made by the bisector of ZB. 

7. The sides of a triangle are 6, 8, and 10. Find the seg¬ 
ments of the sides made by the bisectors of the angles. 

8 . In an inscribed quadrilateral the product of the 
diagonals is equal to the sum of the products 
of the opposite sides. 

Construct DX, making ZXDC= ZADB. Then 
AABD and XCD are similar; and A BCD and AXD 
are also similar. 

9. Given the chords AB and AC from any 
point A on a circle, and AB, a diameter. If the tangent at 
B intersects AB and AC at B and F, and if the chord BC 
is drawn, then the A ABC and AEF are similar. 




§§229,230 FUNDAMENTAL CONSTRUCTIONS 


185 


II. Fundamental Constructions 
Proposition 14. Dividing a Line 

229. Problem. Divide a given line segment into parts 
proportional to any number of given line segments. 



Given the line segments AB^ m, n, and p. 

Required to divide AB into parts proportional to m, n, 
and p. 

Construction. From A draw AX, making any convenient 


Z with AB. 


Post. 1 

On AX, using dividers, take AM=m, MN= 

■n, and NP=p. 

Draw 

BP. 

Post. 1 

At N construct 

NN' II to PB, 


and at M construct 

MM'W to PB. 

§107 


Then M'and AT'are the required points of division. 

The proof is left for the student. It should be observed that § 113 is 
a special case of this problem. 

230. Fourth Proportional. The fourth term of a proportion 
is called the fourth proportional to the terms taken in order. 

Thus, in the proportion ciib = cid, the term d is the fourth propor¬ 
tional to a, b, and c. 












186 


FUNDAMENTAL CONSTRUCTIONS book ill 


Proposition 15. Fourth Proportional 

231. Problem. Construct the fourth 'proportional to 
three given line segments. 



Given the three line segments m, n, and p. 


Required to find the fourth proportional to m, and p. 

Construction. Draw two lines AX and AY forming any 
convenient Z YAX. Post. 1 

Any acute Z will be convenient, although an obtuse Z may be used. 


On AX, using dividers, take AB = m and BC = n. 
Similarly, on AF take AD = p. 


Draw BD. Post. 1 

At C construct a line II to BD, , § 107 

and designate the point where it meets AY as E. 

Then DE is the required fourth proportional. 
AB_AD 
BC DE 

If through two sides of a Aa line is constructed II to the third 
. side, it divides the two sides proportionally. 


Proof. 


§201 


Substituting m, n, p for their equals, AB, BC, AD, we have 


n DE 

/, DE is the fourth proportional to m, n, p. 


Ax. 5 
§230 











187 


§§231,232 CONSTRUCTING PROPORTIONALS 


Proposition 16. Mean Proportional 

232. Problem. Construct the mean proportional be¬ 
tween two given line segments. 



Given the two line segments m and n. 

Required to construct the mean proportional between 
m and n. 


Construction. Draw any convenient line AX. Post. 1 
On AXy using dividers, take 

AC — m and CB = n. 

Bisect AB as at O. § 102 


With 0 as center and OA as radius, construct a semicircle 
as shown. Post. 4 

At C construct the J_ CP, meeting the O at P. § 104 
Then CP is the mean proportional between m and n. 


Proof. 


AQ^QE 

CP CB 


§217 


Substituting m, n for their equals, AC, CB, we have 
m ^CP 
CP n 

.'. CPis the mean proportional between m and n. § 197 


Ax. 5 












188 


FUNDAMENTAL CONSTRUCTIONS book hi 


Proposition 17. Similar Polygons 

233. Problem. Upon a given line segment corresponding 
to a given side of a given polygon construct a polygon 
similar to the given polygon. 



Given the line segment A'B' and the polygon ABCDE. 

Required to construct on A'B\ corresponding to AB, a 
polygon similar to the polygon ABCDE. 

Construction. From A draw the diagonals AC, AD. Post. 1 
From A' construct A'X, A'F, and A% making 

y’=y, 

and z'=z. § 106 

Similarly, from B' construct B'C'y making Z^' = ZR; 
from C' construct C'D\ making /ID’C'A’ — Z^DCA^; and 
from D' construct D'E\ making AE'D'A' = Z.EDA. 

Then A'B'C'D'E’ is the required polygon. 

Proof. AA’B’C' is similar to AABC, 

AA'C'D' is similar to AACD, 

and AA’D'E' is similar to AADE. § 209 

. *. the two polygons are similar. § 226 










EXERCISES 


189 


§ 233 


Exercises. Constructions 

1. If a and two given lines, construct a line equal 
to X where x = y/ah. Consider the special case of a = 8, 6 = 2. 

2. Construct the third proportional to two given line 
segments. 

This means, given two line segments a and 6, find x such that 
a:b = b:x; that is, find a fourth proportional to a, 6, and b. 

3. In Ex. 2 find x both by geometric construction and 
arithmetically when a = 8 in. and 6 = 6 in. 

4. Determine both by geometric construction and arith¬ 
metically the fourth proportional to lines which are 2j in., 
4 in., and 4^ in. long respectively. 

5. Determine both by geometric construction and arith¬ 
metically the mean proportional between lines which are 
2.4 in. and 3.4 in. long respectively. 

6. Find Vt by geometric construction. Measure the 
line and thus determine the approximate arithmetic value. 

7. A map is drawn to the scale of 1 in. to 100 mi. How 
far apart are two places that are 3i in. apart on the map ? 

8. Through a given point P within a given circle 
construct a chord AB such that the ratio 
AP'.BP shall equal a given ratio m:n. 

Construct OPC so that OP:PC=n:m. Then con¬ 
struct CA equal to the fourth proportional to n, m, and 
the radius of the circle. 

9. Given the perimeter, construct a triangle similar to 
a given triangle. 

10. Construct two circles of radii i in. and J in. respec¬ 
tively which shall be tangent externally, and construct a 
third circle of radius 1 in. which shall be tangent to each 
of these two circles and inclose both of them. 




190 


FUNDAMENTAL CONSTRUCTIONS 


BOOK III 


11. Given a line segment 3.5 in. long, divide it both 
internally and externally in the ratio 3:4. 


If AB is the given segment and P' the external point of division 
(§200), then AP' : P'B = 3:4. But AP' = P'B-AB, 
and hence it is possible to compute the length of P'B. 

12. Through a given point P in the arc ^ 
of the chord AB construct a -chord which ^ 
shall be bisected by AB. 



In the figure CD = CP and DE is II to BA. 

B 

13. Through a given external point 
P construct a secant PAB to a given p 
circle so that the ratio PA:AB shall 
equal a given ratio m: n. 

On the tangent PC construct D such that 
PD: DC — m\n. Then construct PA such that PA : PC = PC: PB. Con¬ 
sider any impossible case. 



14. Through a given external point P construct a secant 
PAB to a given circle so that AB = PA • PB, 

If PC is the tangent from P, then PB:PC = 

PC-.PA, or PC^ = PA • PB. But it is required 
that AB^ — PA-PB. What is the relation of 
AB to PC? What is the locus of the midpoints 
of equal chords of a circle ? By constructing a 
tangent, how can you construct the secant PAB 
so that AB = PC ? 





15. Through one of the points of intersection of two 
circles construct a secant such that the 
two chords that are formed shall be in 
a given ratio m : n. 

If X is constructed on the line of centers 
so that OX : XO' = m:n, if MPN is ± to PX, 
and if perpendiculars are drawn from O and O' to MP and PN, what 
follows as to the relation of ilfP to PAT? 





§234 


NUMERICAL RELATIONS 


191 


III. Numerical Relations 
Proposition 18. Side opposite an Acute Angle 

234. Theorem. The square of the side opposite an acute 
angle of any triangle is equal to the sum of the squares 
of the other two sides diminished by twice the product 
of one of those sides and the projection of the other side 
upon it. 



Given the A ARC with an acute ZA, and and b\ the pro¬ 
jections of a and b respectively upon c. 

Prove that 2 Vc. ^ 

Proof. Depending on whether D is between A and B or 


not, we have or a'=6' —c. §5 

Squaring, a'^= 6'^+ 2 b'c. Ax. 6 

Adding to each side of this equation, we have 

h^-^ 6'^+ 2 6'c. Ax. 1 

But h^-\- a'^= a\ and b'^ = b\ § 218 

Substituting (P and b^ for their equals in the above equa- 
tion, we have 2 j Ax. 5 


Pages 191-198, which illustrate the application of algebra to geom¬ 
etry, may be omitted without destroying the sequence. 


















192 


NUMERICAL RELATIONS 


BOOK III 


Proposition 19. Side opposite the Obtuse Angle 

235. Theorem. The square of the side opposite the obtuse 
angle of any obtuse triangle is equal to the sum of the 
squares of the other two sides increased by twice the prod¬ 
uct of one of those sides and the projection of the other 
side upon it 



Given the obtuse A ABC with the obtuse ZA, and a' and b'y 
the projections of a and b respectively upon c. 


Prove that 



Proof. 

a'=6'+c. 

Ax. 10 

Squaring, 

a'"=6'2+c"-|-26'c. 

Ax. 6 

Adding h^ to each side of this equation, we have 




Ax.l 

But 



and 


§218 


Substituting and b^ for their equals in the above 
equation, we have a^^b^+c‘+2b'c. Ax. 5 

The student should notice that if b swings about A so that /.A 
becomes a rt. Z, then b' becomes 0, and hence a'^ = b^ + c^; in other 
words, we have §218. If A A becomes acute, then 6'passes through 
0 and becomes negative, and hence we have § 234. 













§§ 235, 236 


TRIANGLES 


193 


Proposition 20. Squares of Two Sides 

236. Theorem. The sum of the squares of two sides of 
a triangle is equal to twice the square of half the third 
side, increased by twice the square of the median upon it. 

The difference between the squares of two sides of a 
triangle is equal to twice the product of the third side 
and the projection of the median upon it 



Given the A ABC with 6> a, the median m (or CM), and the 
projection m! of m upon the side c. 

Prove that 6^+ a^= 2 AM^ + 2 

and that b‘^—a^=2 cm'. 

Proof. Z.CMA is obtuse, and /LCMB is acute. §§ 124,18 
Since it is given that 6 > a, M lies between A and D. § 118 


Then V = AM‘+ m‘+2AM-m', § 235 

and a^=MB'‘+m^-2MB- m’. §234 

Since MB = AM{% 132), if we add these equals, we have 
6^+a^=2 AM^+2ml Ax. 1 

Subtracting the second equation from the first, we have 
^ 2 MB) = 2 cm'. Ax. 2 


The student should also consider the proposition when a = b. This 
theorem enables us to compute the medians when the three sides 
are known. 











194 


NUMERICAL RELATIONS 


BOOK III 


Exercises. Numerical Relations 


1. Assuming that the area of a triangle is half the prod¬ 
uct of its base and height, as will be proved later, find the 
area of a triangle in terms of its sides. 

At least one of the zi A and B of the A ABC 
is acute. Suppose that Z.A is acute. 

In the A ADC, = b‘^-AD\ § 218, Ax. 2 

In the A ABC, a^ = b‘^ + c^-2c -AD. § 234 

6-2_^c2-a2 


Then 


Hence 


AD 


h‘^ = b‘^- 


2c 

( 62 ^c 2 -( 



4c^ 

46V-(62+c^-a^)^ 

4c^ 

(2 6c+62 + c2- 


c'^ + a^) 


_ _ a‘-^)(2 _ 

4c‘-^ 

^ [jb + cf - a"][a^ - (6 - c)^] 

4c''^ 

_ (a + 6 + c) (6 + c — a) (a + 6 — c) (a — 6 + c) 

4c^ 

Let a + 6 + c= 2s, where s stands for semiperimeter. 

Then • fe + c —a = a + 6 + c —2a = 2s —2a = 2(s —a). 
Similarly, a + 6 — c = 2 (s — c), 

and a —6 + c =2(s —6). 

Hence ■ 2(s-a) • 2(s - 6) ■ 2(a-c)_ 

4c2 


Simplifying, and finding the square root, we have 

2 _ 

h = - Vs(s — a) (s — b) (s — c). 
c 

Hence area of AABC= \ch = ^sis — a)is — b){s — c). 

This proposition dealing with area is included here on account of its 
relation to the numerical theorems given in §§ 234-236. The subject of 
area will be treated fully in Book IV. Similarly, the propositions of 
§§ 234-238 are sometimes given in Book IV and stated in relation to 
the squares on the lines instead of the squares of the lines as here given. 














§236 


EXERCISES 


195 


2. Find the area of the triangle with sides 3 in., 4 in., 5 in. 

Do this by substituting in the formula of Ex. 1, and check by the 
familiar rule that the area is half the product of the base and height. 

3. Using Ex. 1, find to the nearest 0.01 sq. in. the area 
of the triangle whose sides are 21 in., 3 in., 4 in. 

4. Find to the nearest 0.01 in. the diagonal of the square 
of which the side is 7 in. 

5. Find to the nearest 0.01 in. the side of the square of 
which the diagonal is 1 ft. 8 in. 

6. The minute hand and hour hand of a clock are 3 in. 
and 2| in. long respectively. How far apart are the ends of 
the hands at 3 o’clock ? 

7. From a point in the ceiling of a room 12 ft. high 
wires are stretched to two points on the fioor 6 ft. and 10 ft. 
respectively from a point directly beneath the one in the 
ceiling. Find to the nearest 0.01 ft. the lengths of the wires. 

8. The sum of the squares of the segments of two 
perpendicular chords of a circle is equal to the square of 
the diameter. 

If AB, CD are the chords, draw the diameter BEy and draw ACy 
ED, BD. Prove that AC=ED. 

9. The difference between the squares of two sides of a 
triangle is equal to the difference between the squares of 
the segments of the third side made by the perpendicular 
to this side from the opposite vertex. 

10. The square of one of the equal sides of an isosceles 
triangle is equal to the square of any line drawn from the 
vertex to the base, increased by the product of the segments 
of the base. 

11. The three sides of a triangle are 3 in., 4 in., 5 in. Find 
to the nearest 0.01 in. the length of any median. 


196 


NUMERICAL RELATIONS 


BOOK III 


Proposition 21. Bisector of an Angle 

237. Theorem. The square of the bisector of an angle 
of a triangle is equal to the product of the sides which 
form this angle diminished by the product of the seg¬ 
ments made by the bisector upon the third side of the 
triangle. _ 



Given the segment d bisecting AC of the A ABC and forming 
the segments s and s' on AB. 

Prove that d‘^=ab — ss'. 

Proof. Circumscribe a O about A ABC (§ 188), produce 
CM to cut the O as at Q (Post. 2), and draw QB(Post. 1). 
Since ic = a;' (§ 11) and since y = y' 172), we see that 


ABCQ is similar to AMCA, § 209 

Hence CQ:b = a:d; § 205 

whence ab = CQ' d = id-{-MQ)d = d‘^+MQ • d. §198,1 

But MQ-d = ss\ §220 

and hence ab = ss'. Ax. 5 

or d^=ab — ss'. Ax. 2 


This theorem combined with that of § 227 enables us to compute the 
bisectors of the angles of a triangle terminated by the opposite sides, 
if the three sides are known. 








§§ 237, 238 


TRIANGLES 


197 


Proposition 22. Product of Two Sides 

238. Theorem. The product of two sides of any triangle 
is equal to the product of the diameter of the circum¬ 
scribed circle and the altitude upon the third side. 



Given the AARC with the altitude CP (or /z), and CD (or d) 
the diameter of the circumscribed O. 


Prove that 

II 


Proof. Draw 

BD. 

Post. 1 

Then 

/.CPA is a rt. Z, 

§74 

and 

/CBD is a rt. Z. 

§173 

Further, 

X is measured by \ arc BC, 


and 

x' is measured by J arc RC, 

§172 

and hence 

x = x\ 

Ax. 5 


AAPC is similar to ADBC. 

§210 

Hence 

II 

§206 

and 

II 

§ 198,1 


This proposition closes the list of propositions of a semialgebraic 
nature in Book III. As stated on page 191, they may be omitted without 
destroying the geometric sequence. They are needed for the exercises 
on page 198, but not for those on pages 199 and 200. 









198 


NUMERICAL RELATIONS 


BOOK III 


Exercises. Numerical Relations 


If Ex, ly page 19Jf, has been solved, find the areas to the 
nearest 0.01 of triangles with sides as follows: 

1. 4, 5, 6. 2. 6, 8, 10. 3. 7, 8, 11. 4. 1.2, 3, 2.1. 

5. In terms of the sides of a given inscribed triangle, 
find the radius of a circle. 

Consider this exercise only in case §238 and 
Ex. 1, page 194, have been studied. 

Let CD be a diameter. By §238, what do we 
know about thQ products CA • BC and CD' CPI 
What does this tell us of ah and 2 r • CP, where r 
is the radius ? From Ex. 1, page 194, what does CP 
equal in terms of the sides? From the above reasoning show that 

abc 

r = — ; -: 

4 Vs (s — a) (s — 6) (s — c) 



If Ex. 5 has been solved, compute the radii to the nearest 
0.01 of the circles circumscribed about triangles with sides 
as follows: 

6. 3, 4, 5. 7. 27, 36, 45. 8. 7, 9, 11. 9. 10, 11, 12. 


10. Find the medians of a triangle in terms of its sides. 


Omit if § 236 has not been studied. What 
do we know about a^ + 6^ as compared with 
2m2+2(^c)2? 

From this relation show that for the me¬ 
dian m in this figure, 

m = I V2(a2-l-6^) —c^. 



If Ex. 10 has been solved, find to the nearest 0.01 the three 
medians of triangles with sides as follows: 

11. 3, 4, 5. 12. 6, 8, 10. 13. 6, 7, 8. 14. 7, 9, 11. 

15. Find the altitude of a triangle of which the base is 
4 in. and the other sides are 3 in. and 2.5 in. respectively. 









REVIEW EXERCISES 


199 


Exercises. Review 

1. Omitting §§ 234-238, make a list of the numbered 
propositions in Book III, stating under each the proposi¬ 
tions in Books I-III upon which it depends either directly 
or indirectly. 

2. Omitting §§234-238, make another list of the num¬ 
bered propositions, stating under each the propositions in 
Book III which depend upon it. 

3. The tangents to two intersecting circles, constructed 
from any point in their common chord produced, are equal. 

4. The common chord of two intersecting circles, if pro¬ 
duced, bisects their common tangents. 

5. If two circles are tangent externally, the common 
internal tangent bisects the two common external tangents. 

6. If three circles intersect one another, the common 
chords pass through the same point. 

Let two of the chords, AB and CD, meet 
at O. Join the point of intersection E to O, 
and suppose that EO produced meets its two 
circles at two different points P and Q. Then 
prove that OP=OQ(§ 220), and hence that 
the points P and Q coincide. 

7. If the bisector of an exterior angle of a triangle meets 
the opposite side produced, the square of this segment 
of the bisector is equal to the prod- e^^ 

uct of the segments determined by jc 

it upon the opposite side, dimin¬ 
ished by the product of the other two 
sides of the triangle. 

In proving that PC, let CP bisect the exterior 

ZBCX of the ^ABC. Then prove that A ADC and EBC are similar 
(§ 209), and apply § 223. 

PS 





200 


NUMERICAL RELATIONS 


BOOK III 


8. If the line of centers of two circles meets the circles 
at the consecutive points A, B, C, A and meets the common 
external tangent at P, then PA-PD =PB-PC. 

9. The line of centers of two circles meets the common 
external tangent at P, and a secant is drawn from P, cutting 
the circles at the consecutive points W, X, Y, Z. Prove that 
PW-PZ=PX-PY. 

Draw radii to the points of contact, and to W, X, Y, Z. Construct 
perpendiculars upon PZ from the centers of the circles. 

10. In a circle with a radius of 6 in., chords are drawn 
through a point 2 in. from the center. What is the product 
of the segments of each of these chords ? 

11. The chord AB is 6 in. long and is produced through 
B to the point P so that PB = 24 in. Find the length of the 
tangent to the circle from P. 

12. Two line segments AB and CD intersect at O. How 
would you ascertain, by measuring OA, OP, 00, and OP, 
whether the four points A, P, O, and P lie on the same circle ? 

13. This figure shows a center square, an instrument for 
finding the centers of circular objects. The moveable head 
which has the arms OA and OP 
can be fixed by a set screw on 
the blade 00, which always bi¬ 
sects the ZPOA. Show that, if 
OA and OP rest on a circle, 00 
passes through the center, and 
that by placing the square in 
two positions the center of the circle can be determined. 

14. If three circles are tangent externally each to the 
other two, the tangents at their points of contact pass 
through the center of the circle inscribed in the triangle 
formed by joining the centers of the three given circles. 










BOOK IV 

AREAS OF POLYGONS 
1. Fundamental Theorems 

239. Area. If a rectangular piece of paper is 6 in. long 
and 4 in. wide, we may represent the rectangle by the 
figure here shown. We then see that there are 4 small 
squares in each column and that 
there are 6 columns; hence there 
are 6X4 small squares in the 
whole rectangle. Each of these 
squares is 1 in. on a side, and we 
define the area of such a square 
as one square inch (1 sq. in.). 

Each of the small squares is called a unit of area. The 
area of the piece of paper is 6 X 4 sq. in., or 24 sq. in. 

As with the unit of length, a precise definition of these terms for 
the purposes of proof is unnecessary. Among the common units of 
area are 1 sq. in., 1 sq. ft., and 1 sq. mi. Sometimes a unit is taken that 
is not commonly in the form of a square, as in the case of the acre; 
but this measure contains 160 sq. rd., so that the fundamental unit 
in this case is 1 sq. rd. 

In case the sides of a rectangle are considered as incom¬ 
mensurable (§ 164), the subject of areas requires special 
treatment in a manner similar to that used in § 517. For 
the present we shall consider the line segments used in 
Book IV as commensurable, as they are for all practical 
purposes of measurement. 

PS 


-i-1- 

1 1 

1 ! 


1 

1 

1 

j 1 

i 1 


1 

1 

1 

1 ■ 

1 

-1 i 


-1 

1 

L 

-...i 1 , 


1 

1 

1 

1 


201 









202 


FUNDAMENTAL THEOREMS 


BOOK IV 


240. Equivalent Figures. Figures that have equal areas 
are called equivalent figures. 

For example, the figures shown below are equivalent, 
since the area of each figure is 24 times the area of one 
of the small squares of the coordinate paper. 



Since congruent figures can be made to coincide, such 
figures are manifestly equivalent. Equivalent figures can¬ 
not usually be made to coincide, however, and hence they 
are not usually congruent, as is seen above. 



Of the above rectangles, A and B are both congruent and 
equivalent; B and C are equivalent but not congruent, and 
similarly for A and C. 

Since the word "congruent” means identically equal, the word 
'' equal ’ ’ is commonly used to mean equivalent. Thus, since their 
areas are equal, equivalent figures are frequently spoken of as equal 
figures. The symbol = may be used both for "equivalent” and for 
" congruent,” as the conditions under which it is used will determine 
which meaning is to be assigned to it. 

In propositions relating to areas the word " rectangle ” is commonly 
used for area of the rectangle, and similarly for other plane figures. 
It is also the custom to speak of the product of two line segments 
when we mean the product of their numerical measures. 
























































































§§ 240-242 


AREA OF A RECTANGLE 


203 


24L Area of a Rectangle. From the preceding discussion 
we may assume as true the statement that 

The area of a rectangle is the product of the base and the 
altitude. 

In case the sides have a common unit of measure, this is readily 
proved from the figure of § 239. In case they have no common measure, 
the proof is similar to the one given in § 517. 

If the base is 3 in. and the altitude 2 in., the area is 3 X 2 sq. in., or 
6 sq. in. This is the meaning of the expression '' the product of the base 
and the altitude.’^ 

In industrial work 2.' is used for 2 ft. and 2" for 2 in. 


If R stands for the number of units of area of a rectangle 
of base b units and altitude h units, the above statement 
may be written 

R = bh; - 

T> 

whence 6 = -r-» R h 

h 

A h ^ ^ 

and ^~'b' 

formulas that we sometimes use in measuring rectangles. 


242. Ratio of Two Rectangles. In considering the areas of 
two rectangles R and R' we see that 


R _bh 
R' b'h'^ 

Then if h' = h, we have 
R^bh^b, 
R' b'h 6" 




that is, rectangles with equal altitudes are to each other as 
their bases. 

Similarly, rectangles with equal bases are to each other as 
their altitudes. 

As stated in §240, the word rectangles ” is here used for the 
areas of rectangles. 

PS 





204 


FUNDAMENTAL THEOREMS 


BOOK IV 


Exercises. Areas of Rectangles 

1. Find the ratio of a lot 180 ft. long and 120 ft. wide 
to a field 80 rd. long and 40 rd. wide. 

2. A square and a rectangle have equal perimeters of 
576 in., and the length of the rectangle is five times the 
width. Which has the greater area ? How much greater ? 

3. On a certain map the linear scale is 1 in.=20 mi. 
How many acres are represented by a square | in. on a side ? 

4. Find the area of a gravel walk 7 ft. wide which sur¬ 
rounds a rectangular plot of grass 80 ft. long and 50 ft. wide. 

5. Find the number of rods in the perimeter of a square 

field that contains exactly an acre. -jr~l 

6. Find the number of square inches 

in the cross section of this L beam. 4 J 

7. A machine for planing iron plates 
planes a surface 1 in. wide and 9 ft. long 
in 1 min. At the same rate per square 
inch, how long does it take to plane a 
plate 12 ft. long and 6 in. wide, allowing 
28min. for adjusting the machine during the process? 

8. How many tiles, each 6 in. square, does it take to 
cover a floor 36 ft. 6 in. long by 18 ft. wide ? 

9. The length of a rectangle is four times the width. 
If the perimeter is 120 ft, what is the area ? 

10. Along two adjacent sides of a rectangular field 120 rd. 
long and 80 rd. wide a road 4 rd. wide is laid out inside the 
field. How many acres are taken for the road ? 

11. From one end of a rectangular sheet of iron 12 in. 
long a square piece is cut off such that it leaves 36 sq. in. 
in the rest of the sheet How wide is the sheet ? 












§243 


AREA OF A PARALLELOGRAM 


205 


Proposition 1. Area of a Parallelogram 

243. Theorem. The area of a parallelogram is the 
product of the base and the altitude. 



Given the OJABCD with base h and altitude h. 

Prove that the area of OJABCD = bh. 

Proof. At B construct BX J_ to CD, or CD produced, and 


at A construct AY -L to CD, or CD produced. § 105 

The only cases which require special attention are shown above. 

Then ArislltoSX. §57 

ABXY is a □ with base 6 and altitude h. § 72 
Since AY=BX and AD = BC, § 78 

then rt. AADY is congruent to rt. A BOX. § 71 

Now, considering the quadrilateral ABCY, we have 
ABCY- A BCX = □ ABXY, 

and ABCY~AADY=nABCD. Ax. 10 

But ABCY-ABCX=ABCY- AADY, Ax. 2 

and hence aABXY^^CJABCD. Ax. 5 

But C]ABXY=bh. §241 

.•.OABCD = hh. Ax. 5 


By this theorem we have proved the correctness of a formula with 
which the student has long been familiar. 


















206 


FUNDAMENTAL THEOREMS 


BOOK IV 


Proposition 2. Area of a Triangle 

244. Theorem. The area of a triangle is half the product 
of the base and the altitude. 



Given the A ABC with base b and altitude h. 

Prove that the area of AA5C= I hh. 

Proof. With CA and AB as adjacent sides construct the 


OABDC. § 107 

Then /\ABC = { OABDC. § 77 

But OABDC =hh. §243 

.\AABC=lhh. Ax. 4 


245. Corollary. Triangles with equal bases and equal alti¬ 
tudes are equivalent; and similarly for parallelograms. 

For, whatever the shape, the area of the A is ^ hh, and the area of 
the EJ is hh. 

246. Corollary. Triangles with equal bases are to each 
other as their altitudes; triangles with equal altitudes are 
to each other as their bases; any two triangles are to each 
other as the products of their bases and altitudes; and 
similarly for parallelograms. 

Has this been proved for E] ? What is the relation of a A to a d 
of equal base and equal altitude? What must then be the relations 
of A to one another ? Can the same be proved for [U ? 










§§ 244-248 TRIANGLE AND TRAPEZOID 207 

Proposition 3. Area of a Trapezoid 

247. Theorem. The area of a trapezoid is half the 
product of the altitude and the sum of the bases. 



Given the trapezoid ABCD with bases h and b' and altitude h. 
Prove that the area of ABCD = J (6 -f 6'). 


Proof. Draw 

the diagonal AC. 

Post. 1 

Then 

AABC=i~bh, 


and 

AACD = l-b'h. 

§244 

Hence 

ABCD=^lbh + \b'h‘, 

Ax. 1 

that is, 

ABCD = lh{b + b'). 



248. Area of an Irregular Polygon. The area of an irregular 
polygon may be found by dividing the polygon > 
into triangles and trapezoids and then find- ^ 

ing the area of each of these triangles and r' \ 

trapezoids separately. -4 

A common method used in land surveying is as U-- 

follows : Draw the longest diagonal, construct perpen- \. -y 

diculars upon this diagonal from the other vertices of 

the polygon, as shown in the figure, and then measure 

each of the dotted lines. The sum of the areas of the right triangles, 

rectangles, and trapezoids thus formed is the area of the polygon. 

The student should see that he can now measure any rectilinear figure. 















208 


FUNDAMENTAL THEOREMS 


BOOK IV 


Exercises. Areas 


1. Find the area of a trapezoid of which the bases are 
17 in. and 13 in. respectively and the altitude is 7.5 in. 

2. A railway embankment is 30 ft. 
high, 80 ft. wide at the top, and 116 ft. 
wide at the bottom. Find the area of 
the cross section. 

3. A canal is 36 ft. deep, 240 ft. wide at the top, and 
200 ft. wide at the bottom. Find the area of the cross section. 




4. A polygon of six sides is made up of six congruent 
triangles such that the base of each triangle is 4 in. and 
its altitude is 2V3in. Find the area of the polygon to the 
nearest 0.1 sq. in. 

5. In surveying the field here shown a 
surveyor laid off a north-and-south line NS 
through A and then found that = 6 rd., 

CC"=10 rd., DD'= 7 rd., R'A=8rd., B'C'==12 rd., 

C'D'= 4rd. Find the area of the field. 

6. In Ex. 5, what would be the area if each 
of the given measurements were doubled? 

7. The area of a trapezoid is the product of the alti¬ 
tude and the line segment joining the midpoints of the 
nonparallel sides. 

8. Find the area of the cross section of 
the steel girder here shown. 

9. In Ex. 8, what would be the area if 
each of the given measurements were multi¬ 
plied by three ? 

10. The product of the sides forming the* 
right angle of a right triangle is equal to ' 

the product of the hypotenuse and the altitude upon it. 



















§249 


AREAS 


209 


Proposition 4. Ratios of Areas of Triangles 

249. Theorem. If an angle of one triangle is equal to 
an angle of another^ the triangles are to each other as 
the products of the sides forming the equal angles. 



Given the ^ABC and ADE with the common ZA. 


Prove that 


A ABC _ AB -AC 
A ADE AD-AE' 


Proof. Draw 
Then 

and 


BE, 

A ABC _ AC 
A ABE Ae' 
AABE _AB 
A ADE ad' 


Post. 1 


§246 


because A with equal altitudes are to each other as their bases. 


Since we are considering numerical measures, we may 
treat the terms of these proportions as numbers. 

Taking the product of the first members and the product 
of the second members of these equations, we have 

AABE-AABC _ AB-AC 3 

AADE • A ABE AD-AE' 


Then, canceling AABE, we have the proportion 
AABC _ AB -AC 
AADE AD-AE' 
















210 


FUNDAMENTAL THEOREMS 


BOOK IV 


Proposition 5. Similar Triangles 

250. Theorem. The areas of two similar triangles are to 
each other as the squares on any two corresponding sides. 



Given the similar AABC and A'B'C', 
Prove that 


AABC _ 
AA'B'C' 


Proof. Since AABC is similar to AA'B'C\ 
we have ZA = ZA'. 

AABC _ he 

■ 7 f / ’ 

be 


Then 


AA'B'C' 

because • • • the A are to each other as the products of the 
sides forming the equal A ; 


Given 

§205 

§249 


that is, 
But 


AABC b c 


AA’B'C' b' c' 

— = — 
b'~c'' 


or 


Substituting for its equal, ^»we have 
c 0 

AABC _£ £ 
AA'B'C ~ c'’ c'* 
AABC __ 

AA'B'C c'"’ 


§205 


Ax. 5 














§§ 250 , 251 


AREAS OF POLYGONS 


211 


Proposition 6. Areas of Polygons 

251. Theorem. The areas of two similar polygons are to 
each other as the squares on any two corresponding sides. 



Given two similar polygons with areas S and S' respectively. 

Prove that * 

S' a'" 

Proof. By drawing all the diagonals from any two corre¬ 
sponding vertices the two similar polygons are separated 


into the similar AP, P'; Q, Q'; P, R\ 

§225 

Then 

^ 6^“ AP a" 

AP' c'“ AQ' b'^~AP' a'^‘ 

§250 

Hence 

AR _ AQ AP 

AR' AQ’ AP’’ 

Ax. 5 

Then 

AR+AQ+AP AP. 
AP'+AQ'+AP' AP' 

§ 198, 8 

But 

AP a\ 

AP' a"‘ 

Proved 

Then 

AR+AQ+AP 
AP'+AQ'+AP' a'"' 

Ax. 5 

and hence 

S_ a" 

S' a'"’ 

Ax. 10 












212 


FUNDAMENTAL THEOREMS' 


BOOK IV 


Proposition 7. Pythagorean Theorem 

252. Theorem. The square on the hypotenuse of a right 
triangle is equivalent to the sum of the squares on the 
other two sides. 



Given the rt. A ABC with the rt. Z C, and the squares con¬ 
structed on the sides a, 5, c respectively. 

Prove that c^=a^-\- 

Proof. Construct CX II to AR (§ 107), and draw BQ and CR. 


Since x and z are rt. Z, then XPCB is a st. Z. 

§13 

Hence PCB is a st. line; and similarly for ACN. 

§18 

Then 

AR = AB, AC = AQ, 

§15 

and 

zrac=a:baq. 

Ax.l 


AARC is congruent to AABQ. 

§40 

But 

[DAX=2AARC, 

§244 

because they have the same base AR and the same altitude RX. 

Similarly, 

V=2 AABQ = 2 AARC. 



□^X=6". 

Ax. 5 

Similarly, 




nAX+nBX=b^+a\ OYc^=a^-\-h\ 

Ax.l 












§§ 252-254 


PYTHAGOREAN THEOREM 


213 


253. Corollary. The square on either side of a right tri¬ 
angle is equivalent to the difference between the square on 
the hypotenuse and the square on the other side. 

For, since § 252 

then c^—a^=b‘\ Ax. 2 

254. Pythagorean Theorem. The fact that the square on 
the hypotenuse is equivalent to the sum of the squares on 
the other two sides has, of course, long been known to the 
student. It is usually learned in arithmetic, and we have 
already given an algebraic proof in §218. Various geo¬ 
metric proofs may be given, but the one in § 252 is the 
most satisfactory for beginners. This proof is attributed 
to Euclid, a famous mathematician who lived in Alexandria, 
in Egypt, about 300 B.C. Euclid wrote the first great text¬ 
book on geometry, and taught in the world's first great 
university, an institution founded by one of the Greek 
kings of Egypt. 

It is thought, as stated in § 218, that Pythagoras gave 
the first proof of this theorem about 525 b.c., but it is not 
certain that he did so. Pythagoras founded the world's 
first great school of mathematics at Crotona, in the south¬ 
eastern part of Italy, which was then a Greek colony. 

If § 218 has been thoroughly mastered, § 252 may be omitted. 

From a study of the theorem we see that the diagonal 
and side of a square are incommensurable. 

For = + 

or (P = 2s^. 

Hence d = s V2. 

Since V2 may be carried to as many decimal places 
as we please, but cannot be exactly expressed as a rational number, it 

has no common measure with 1. That is, - = V2, an incommensurable 
number, and hence the diagonal and side are incommensurable. 




214 


FUNDAMENTAL THEOREMS 


BOOK IV 


Exercises. Areas 

Find the areas of the 'parallelograms 'whose bases and 
altitudes are respectivel'y as follows: 

1. 4.5 in., 2f in. 2. 5.4 ft, 2.4 ft 3. 4 ft 6 in., 14 in. 

Find the areas of the triangles whose bases and altitudes 
are respectivel'y as follows: 

4. 2.8 in., 3 in. 5. 13 ft, 6 ft 6. 4 ft 6 in., 2 ft 

Find the areas of the trapezoids whose bases are the first 
two of the following numbers, and whose altitudes are the 
third numbers: 

7. 2|ft, lift; 5in. 8. 4ft 7in., 3 ft; 16in. 

Fmd the altitudes of the parallelograms whose areas and 
bases are respectively as follows: 

9. 20sq.in., lOin. 10. 8 sq. ft, 3 ft 11. 7 sq. ft, 2 ft 

Fi'ud the altitudes of the triangles whose areas and bases 
are respectively as follows: 

12. 9 sq. in., 4 in. 13. 7 sq. ft, 2 ft 14. 11 sq. yd., 3 yd. 

15. Find the altitude of the trapezoid whose area and 
bases are 33 sq. in., 5 in., and 6 in. respectively. 

Given the sides of a right triangle as follows, find the 
hypotenuse to the nearest 0.01 ft.: 

16. 60 ft., 80 ft 17. 40 ft, 60 ft. 18. 7 ft. 6 in., 9 ft. 

Given the hypotenuse and one side of a right triangle as 
follows, find the other side to the nearest 0.01ft.: 

19. 25 ft., 20 ft 20. 20 ft., 12 ft. 21. 3 ft. 4 in., 2 ft 


254 


EXERCISES 


215 



E 




B 


22. The square constructed upon the sum of two line 
segments is equivalent to the sum of the 
squares constructed upon the two segments, 
increased by twice the rectangle of the 
segments. 

Given the two line segments AB and BC, their 
sum AC, and the squares AG and AE constructed 
upon AC and AB respectively. Complete the figure as shown. Then 
the square AG is the sum of the squares AE, EG and the CD DE, CE. 

This proves geometrically the algebraic formula 

(a + 6)2=2 a6 4- 6^. 

23. The square constructed upon the difference between 

F 
E 


B 


two line segments is equivalent to the sum 
of the squares constructed upon the two 
segments, diminished by twice the rectangle 
of the segments. 

Given the two line segments AB and AC, their 
difference BC, the square AF constructed upon AB, 
the square AG upon AC, and the square CE upon 
BC. Complete the figure as shown. Then the square CE is the differ¬ 
ence between the whole figure and the sum of two rectangles. 

This proves geometrically the algebraic formula 

(a — hf = d^—2ah-{- b^. 

24. The difference between the squares constructed upon 
two line segments is equivalent to the rectangle of the sum 


and difference of these lines.’ 

Given the squares AD and CE constructed upon 
AB and BC respectively. Show that the difference 
between the squares AD and CE is equivalent to 
the □ AF, with dimensions AB -f- BC and AB — BC. 

This proves geometrically the algebraic formula 

a‘^-b-‘=(a + b)(a-b). " ^ 

Before our present algebra was invented the algebraic laws given 
in Exs. 22-24 were proved as above by geometry. 


E 





















216 


FUNDAMENTAL THEOREMS 


BOOK IV 


25. An extension ladder 77 ft. long is placed with its top 
against a wall, and its foot 46.2 ft. from the base of the 
wall. How high, to the nearest 0.1 ft, does the ladder 
reach on the wall ? 


Sun 


26. Galileo (1564-1642), who was the first to use the tele¬ 
scope in astronomy, found the height of a mountain on the 
moon by the aid of the Pythagorean Theo¬ 
rem. On a map of the moon he measured the 
distance d from the top of the mountain M 
when it was touched by the sun's rays to the 
line dividing the light half of the moon from 
the dark half. Representing the height by h and the radius 
of the moon by r, he saw that 

{h + rf=7^+d\ 



Find h, given that the radius of the moon is 1081 mi. 

Students who have had quadratics should solve thi s equatio n for h, 
then substitute 1081 for r, and find that /i=—lOSl+VlOSl^+cJ^ where 
h is in miles. An approximate solution in feet is h = 2.44 d^. 


27. Find a formula for the altitude h of an equilateral 
triangle in terms of its side s. 

28. Find a formula for the side s of an equilateral triangle 
in terms of its altitude h. 


29. If A is the area of an equilateral triangle with side 
s, prove that A = ls^ Vs. 

30. Find the length of the longest chord and of the 
shortest chord that can be drawn through a point 1 ft. 
from the center of a circle with a radius of 20 in. 

31. If the diagonals of a quadrilateral intersect at right 
angles, the sum of the squares on one pair of opposite sides 
is equivalent to the sum of the squares on the other pair. 

32. The area of a rhombus is half the product of its 
diagonals. 





§254 


EXERCISES 


217 


33. Two triangles are equivalent if the base of the first 
is equal to half the altitude of the second, and the altitude 
of the first is equal to twice the base of the second. 

34. The area of a circumscribed polygon is half the 
product of the perimeter of the polygon and the radius of 
the inscribed circle. 

35. If equilateral triangles are constructed on the sides of 
a right triangle, the triangle on the hypotenuse is equiva¬ 
lent to the sum of the triangles on the other two sides. 

36. If similar polygons are constructed on the sides of a 
right triangle as corresponding sides, the polygon on the 
hypotenuse is equivalent to the sum of the polygons on the 
other two sides. 

Ex. 36 is one of the general forms of the Pythagorean Theorem. 

37. Every line drawn through the intersection of the 
diagonals of a parallelogram bisects the parallelogram. 

38. If lines are drawn from any point within a parallelo¬ 
gram to the four vertices, the sum of either pair of triangles 
with parallel bases is equivalent to the sum of the other pair. 

39. If a quadrilateral with two sides parallel is bisected 
by either diagonal, the quadrilateral is a parallelogram. 

40. The line that bisects the bases of a trapezoid divides 
the trapezoid into two equivalent parts. 

41. The triangle formed by two lines drawn from the 
midpoint of either of the nonparallel sides of a trapezoid 
to the opposite vertices is equivalent to half the trapezoid. 

42. The sides of a triangle are 1.4 in., 1.2 in., and 1.4 in. 
respectively. Is the largest angle acute, right, or obtuse ? 

43. The sides of a triangle are 9.5 in., 14.1 in., and 17 in. 
respectively. Is the largest angle acute, right, or obtuse ? 

PS 


218 


FUNDAMENTAL THEOREMS 


BOOK IV 


44. Find to the nearest 0.1 sq. in. the area of an isosceles 
triangle whose perimeter is 28 in. and whose base is 8 in. 

45. Upon any two sides AC and BC of a given ^ABC 
the HJ CM and CN are constructed. 

Two sides of these parallelograms 
are produced to meet at P as here 
shown, the line PC is drawn and M. 
produced so that QR = PC, and then 
the CJAT is constructed with BT 
equal to and parallel to QR, Prove that CM-\-CN = AT. 

This interesting generalization of the Pythagorean Theorem is due 
to the Greek geometer Pappus, about A.D. 300. It is not difficult to 
derive the Pythagorean Theorem from it by starting with a right 
triangle and by making CM and CN squares. 

46. Prove the Pythagorean Theorem by 
using this figure. 

Show that the four large right triangles are con¬ 
gruent. If the two triangles marked T and T' are 
taken from the whole figure, there remains the sum 
of the squares on the two sides. If the other two triangles are taken 
from the whole figure, there remains the square on the hypotenuse. 

47. Find the area of a right triangle if the hypotenuse 
is 3.4 in. and one of the other sides is 1.6 in. 

48. Find the ratio of the altitudes of two equal triangles 
if the base of one is 3 in. and that of the other is 9 in. 

49. The bases of a trapezoid are 68 in. and 60 in., and 
the altitude is 4 in. Find the side of a square with the 
same area.^ 

50. The cross section of a V-thread on 
a screw is an equilateral triangle. The 
distance p between successive threads 
is known as the pitch of the thread, and the distance d as the 
depth of the thread. If p = J in., what is the value of d ? 

















§§ 255,256 FUNDAMENTAL CONSTRUCTIONS 


219 


II. Fundamental Constructions 
Proposition 8. Sum of Two Squares 

255. Problem. Construct a square equivalent to the 
sum of two given squares. 



a2 



! 

B 

a 

1 t 

1 

C\ 


a 

b ' C > b Ai 


Given the squares and W with sides a and h respectively. 

Required to construct a square equivalent to a^+ W. 

Construction. On any line construct the rt. ZC(§104), 
and on its arms take CB = a and CA = b. 

Draw AB, or c. Post. 1 

With c as a radius, construct the required square c^ by 
drawing arcs as shown. Post. 4 

Proof. c^=a^-{'b^» §252 

256. Purpose of These Constructions. Since the area of a 
square is easily found, it is often advantageous to trans¬ 
form a rectilinear figure into a square. It is also helpful 
to combine several squares into a single square, by first 
finding a square equivalent to two of the squares, and then 
combining this square with a third one, and so on. 

The student may omit §§ 255-260 without interfering with the sub¬ 
sequent work, and should omit §§ 261 and 262 unless preparing for more 
advanced work in mathematics. In some courses §257 is required. 



















220 


FUNDAMENTAL CONSTRUCTIONS book iv 


Proposition 9. Transforming a Polygon 

257. Problem. Construct a triangle equivalent to a 
given polygon. 



Given the polygon ABCDEF. 

Required to construct a A equivalent to ABCDEF. 


Construction. Let B, C, and D be any three consecutive 
vertices of the polygon. 


Draw 

the diagonal DB. 

Post. 1 

From C construct a line II to DB. 

§107 

Produce 

AB to meet this line at Q, 

Post. 2 

and draw 

DQ. 

Post. 1 


Similarly, draw EQ, and from D construct a line II to EQ, 
meeting AB produced at R, and draw ER. 

Continue to reduce the number of sides of the polygon 
until the required AEPR is obtained. 

Proof. Polygon AQDEF has one side less than ABCDEF. 

Now ABDEF is common to both polygons, 
and ABQD = ABCD. § 245 

.*. AQDEF=ABCDEF. Ax. 1 

Similarly, AREF=AQDEF, 2indEPR = AREF. 








'§§ 257-260 


EQUIVALENT POLYGONS 


221 


Proposition 10. Transforming a Parallelogram 

258. Problem. Construct a square equivalent to a given 
parallelogram. 



Given the EJABCD with base b and altitude h. 

Required to construct a square equivalent to EJABCD. 

Construction. On any line take NP = h and PM — b. 
Construct the mean proportional s to /i and b. § 232 
With s as radius, construct the required square s^ by 


drawing arcs as shown. 

Post. 4 

Proof. Since 

s is _L to NMy 

Const. 

then 

h:s = s:b. 

§217 


.\s^=bh. 

§ 198,1 

But 

OABCD = bh. 

§243 


.\s^ = nABCD. 

Ax. 5 


259. Corollary. Construct a square equivalent to a given 
triangle. 

Construct s so that h:s = s:\h. 

260. Corollary. Construct a square equivalent to a given 
polygon. 

Reduce the polygon to an equivalent A (§ 257), and then construct a 
square equivalent to this A (§ 259). 













222 


FUNDAMENTAL CONSTRUCTIONS book iv 


Exercises. Constructions 

1. Construct a square which shall have twice the area 
of a given square. 

2. Construct a triangle equivalent to the sum of any 
two given triangles. 

3. Construct a right triangle equivalent to a given 
oblique triangle. 

4. Construct a rectangle equivalent to a given parallelo¬ 
gram, and with its altitude equal to a given line. 

5. Construct a triangle equivalent to a given triangle, 
and with one side equal to a given line. 

6. Construct a right triangle equivalent to a given tri¬ 
angle, and with one of the sides of the right angle equal 
to a given line. 

7. Construct a right triangle equivalent to a given tri¬ 
angle, and with its hypotenuse equal to a given line. 

8. Divide a given triangle into two equivalent parts by 
a line through a given point P in the base. 

9. Construct a polygon similar to two given similar 
polygons and equivalent to their sum. 

Exs. 9-12 are often given in older geometries as fundamental con¬ 
structions, but in later textbooks they are usually omitted or are 
given as optional problems. Since they are not needed in proving other 
propositions, they may be omitted except by students who are special¬ 
izing in mathematics. 

10. Construct a polygon similar to a given polygon and 
such that it has a given ratio to it. 

11. Construct a polygon similar to a given polygon and 
equivalent to another given polygon. 

12. Construct a square which shall have a given ratio to 
a given square. 


§261 


SUPPLEMENTARY CONSTRUCTIONS 


223 


III. Supplementary Constructions 
Proposition 11. Constructing a Parallelogram 

261. Problem. Construct a parallelogram equivalent 
to a given square, and with the sum of its base and 
altitude equal to a given line. 



Given the square with side s, and the line AB, 

Required to construct a O equivalent to s\ and with the 
sum of its base and altitude equal to AB. 

Construction. Bisect AB as at O (§ 102), and with O as 
center and OA as radius, construct a semicircle (Post. 4). 

At A construct a ± to AB (§ 104), and on it take AC=s. 

At C construct CD II to ABy cutting the O at P. § 107 

At P construct PQ J- to AB. § 105 

Then any O, as M, with A Q for altitude and QB for base 
is equivalent to 

Proof. Since^Q :PQ=PQ: QB^ 217), then Pq'‘=AQ-QB, 
and since PQ is II to CA (§ 67), we have PQ = CA=s(§ 80). 

.•.AQ-QB=s\ Ax. 6 

Then M=AQ ■ QB= s\ § 243, Ax. 5 

This theorem solves geometrically the equations x^y — a^xy — h. 














224 


SUPPLEMENTARY CONSTRUCTIONS book iv 


Proposition 12. Constructing a Parallelogram 

262. Problem. Construct a parallelogram equivalent to 
a given square^ and with the difference between its base 
and altitude equal to a given line. 



Given the square with side 5, and the line AB. 

Required to construct a O equivalent to s^, with the differ¬ 
ence between its base and altitude equal to AB. 

Construction. Bisect AB as at O (§ 102), and with O as 


center and OA as 

radius, construct a O (Post. 4). 


At A construct 

a tangent to the O, 

§195 

and on it take 

AC=s. 


Through 0 draw CD as shown. 

Post. 1 

Then any O, as 

CM, with CD for its base and CE for its 

altitude, is equivalent to s^. 


Proof. 

CD:s = s: CE. 

§222 


:.s^=CD^CE. 

§ 198,1 

But 

CM= CD • CE. 

§243 


.*. CM=sl 

Ax. 5 

Also, 

CD-CE=ED=AB. 

§ 134, 3 


By this theorem we solve geometrically the algebraic problem of 
finding x and y in the equations x — y = a,xy = b. 













§262 


REVIEW EXERCISES 


225 


Exercises. Review 


1. Omitting §§255-262, make a list of the numbered 
propositions in Book IV, stating under each the proposi¬ 
tions in Books I-IV upon which it depends either directly 
or indirectly. 

2. This figure shows an angle cut by parallel lines. Prove 
that x = ah, and thus show that we may 
construct a line segment equal to the prod¬ 
uct of two line segments. 

We thus see that we may think of a line, as well 
as a rectangle, as representing the product of two line segments. 




3. Draw a figure of about this shape. 

Then construct a triangle equivalent to this 
polygon. Finally, construct a square equiva¬ 
lent to the triangle, measure the square, and 
thus find the area of the original polygon. 

4. Construct a square equivalent to the difference be¬ 
tween two given squares. 

5. This figure represents the cross section of a barn. 
Find the area of the section. 

In finding the number of cubic feet in the barn 
we multiply the area of the cross section by the 
length of the barn. This shows a reason for finding 
the areas of the cross sections of barns, pipes, canals, 
railway embankments, and the like. 

6. In this figure the [JJBCDA and ECDF are equivalent. 
Prove that the triangle formed by join¬ 
ing F to A and B is equivalent to either , 
parallelogram. 

7. In the figure of Ex. 6 draw the ^ ^ 
diagonals AC and FC. Then prove the 
quadrilateral ACFD equivalent to either parallelogram. 














226 


SUPPLEMENTARY CONSTRUCTIONS book iv 




8. In surveying the field ABCD a sur¬ 

veyor runs a north-and-south line through 
A, and from it lays off the CC\ and 

DD\ By measuring he finds that BB'=SS rd., 

CC' = 35 rd., DD' = 14 rd., B'A=2S rd., B'C'= 

42 rd., and AZ)' = 26rd. Find the area of 
the field in square rods; in acres. 

9. Wishing to find the area of a field ABCD bounded 
on one side by a river, a surveyor made a map as here 
shown by constructing the Ji AD, 


P'P, Q’Q, R'R, BC to AB, He found 
that AP' = 26 rd., P'Q' = 18 rd., Q’R' 

= 23rd., i2'5 = 18rd., AL> = 35rd., ^__ 

PP'=42rd., QQ'=32rd.,i?iJ'=38rd., 

^C=35rd. Find the approximate area of the field. 

10. In this figure, ABCD is a parallelogram. Prove that 
APQB is equivalent to APR A, 

11. Generalize Ex. 10 by first let¬ 
ting P move down to rest on the 
line DC and seeing if Ex. 10 holds 
true. Then let P move down below 
DC so as to lie within the parallelo¬ 
gram, and let Q lie on AP produced and R on BP produced. 

12. If P is any point in the diagonal AC of OABCD, then 
AABP is equivalent to AAPD. 

13. A surveyor wishes to divide a field APCD into two 
equivalent parts by a line DP drawn 
from the vertex D. How should he 
proceed to do it ? 

Let M bisect AC and construct MP II to 
DB. From this suggestion show how the 
surveyor solved the problem. 












BOOK V 

REGULAR POLYGONS AND THE CIRCLE 
I. Fundamental Theorems 

263. Regular Polygon. A polygon that is both equiangular 
and equilateral is called a regular polygon (§ 92). 

264. Circumscribed and Inscribed Circles. It will be proved 
in §§ 269 and 270 that a circle can be circumscribed about, 
and a circle can be inscribed in, any regular polygon (§ 156), 
and that these circles are concentric (§ 157). 

265. Radius. The radius of the circle circumscribed about 
a regular polygon is called the radius of 
the polygon. 

In this figure, r is the radius of the polygon. 

266. Apothem. The radius of the circle 
inscribed in a regular polygon is called 
the apothem of the polygon. 

In the figure, a is the apothem of the polygon. The apothem is evi¬ 
dently perpendicular to the side of the regular .polygon (§ 147). 

267. Center. The common center of the circles circum¬ 
scribed about and inscribed in a regular polygon is called 
the center of the polygon. 

268. Angle at the Center. The angle between the radii 
drawn to the extremities of any side of a regular polygon 
is called an angle at the center of the polygon. 

In the figure, m is an angle at the center of the polygon. 

227 






228 


FUNDAMENTAL THEOREMS 


BOOK V 


Proposition 1. Circumscribed Circle 

269. Theorem. A circle can he circumscribed about any 
regular polygon. 



Given the regular polygon ABODE. 

Prove that a O can be circumscribed about ABODE. 

Proof. Let O be the center of a O constructed through 


three vertices A, B, C of the polygon. 

§190 

Draw 

OA, OB, 00, OD. 

Post. 1 

Then 

OB =00. 

§ 134,1 

Further, 

AB=OD. 

§263 

Also, 

ZOBA = ZDOB, 

§263 

and 

ZOBO = ZOOB. 

§42 


/. ZOBA = ZDOO. 

Ax. 2 

Then 

AOAB is congruent toAODO, 

§40 

and hence 

OA = OD. 

§38 

Then the’ O through A, B, G passes through D. 

§ 134,6 


In like manner, it can be proved that the O through B, C, 
and D passes through E\ and so on. 

Hence the O constructed with O as center and OA as 
radius is circumscribed about the polygon. § 156 








§§ 269-273 CIRCLES AND A POLYGON 229 

Proposition 2. Inscribed Circle 

270. Theorem. A circle can he inscribed in any regular 
polygon, ^ _ 



Given the regular polygon P. 

Prove that a O can he inscribed in P. 

Proof. Let 0 be the center of the O circumscribed about 
polygon P. § 269 

Since the sides of P are equal chords of the circum¬ 
scribed O (§ 156), they are equidistant from O. § 150 

Hence the G constructed with O as center and with the 
± OA as radius (§ 146) is inscribed in the polygon. § 156 

271. Corollary. The angles at the center of any regular 
polygon are equal, and each is supplementary to an interior 
angle of the polygon. 

The A at the center are corresponding A of congruent A. 

Further, in the figure of § 269, AAOB + Z OB A + ABAO = 180°, and 
ZBAO = ZCBO. Hence ZAOP + ZCPA =180° 

272. Corollary. An equilateral polygon inscribed in a circle 
is a regular polygon. 

Why are the A also equal ? 

273. Corollary. An equiangular polygon circumscribed 
about a circle is a regular polygon. 

By joining consecutive points of contact of the sides show that cer¬ 
tain isosceles A are congruent, and thus prove the polygon equilateral. 








230 


FUNDAMENTAL THEOREMS 


BOOK V 


Proposition 3. Inscribed and Circumscribed Polygons 

274. Theorem. If a circle is divided into any number 
of equal arcs, the chords of these arcs form a regular 
inscribed polygon; and the tangents at the points of 
division form a regular circumscribed polygon. 



Given the O 0 divided into equal arcs hy A, B, C, D, and E, 
the chords AB, BC, CD, DE, EA, and the tangents PQ, QR, 
RS, ST, TP ditE, A, B, C, D respectively. 

Prove that ABODE is a regular inscribed polygon and 
that PQRST is a regular circumscribed polygon. 

Proof. The arcs AB, BC, CD, DE, EA are equal. Given 


Hence AB=BC=CD = DE = EA, § 139 

because if two arcs • • • are equal, the arcs have equal chords. 

Also, ABODE is inscribed in the O. § 156 

.*. ABODE is a regular inscribed polygon. § 272 
Since the arcs are equal, Given 

ZP=ZQ = ZR = ZS=ZT, § 179 

because an /.formed by two tangents •••is measured by half 
the difference between its intercepted arcs. 

Also, PQRST is circumscribed about the O. § 156 


PQRST is a regular circumscribed polygon. § 273 









§§ 274-277 


POLYGONS AND A CIRCLE 


231 


275. Corollary. Tangents to a circle at the vertices of a 
regular inscribed polygon form a regular circumscribed 
polygon of the same number of sides. 

For it is shown in §274 that PQRST is a regular circumscribed 
polygon. It has as many sides as there are vertices of ABCDE, and 
ABODE has as many vertices as it has sides. 

276. Corollary. Tangents to a circle at the midpoints of 
the arcs of the sides of a regular inscribed polygon form a 
regular circumscribed polygon, whose sides 
are parallel to the sides of the inscribed poly¬ 
gon and whose vertices lie on radii produced 
of the inscribed polygon. 

Since M is the midpoint of arc AB (given), then 
ZA'OM = AB'OM (§ 137). Also, OA = OB (§ 134,1), 
and OM is a common side. Hence Oilf bisects AB (§§ 40,38). Then, since 
two corresponding sides AB and A'B' are both ± to OM (§§ 142, 147), 
they are II (§ 57). 

Further, since the tangents MB' and NB' intersect at a point equi¬ 
distant from OM and ON (§ 149), they intersect upon the bisector of 
ZMON (§ 183). But OB bisects ZMON (§ 137). Hence MB' and NB' 
intersect on OB produced. 

277. Corollary. Lines drawn from each vertex of a regular 

inscribed polygon to the midpoints of the arcs of adjacent 
sides of the polygon form a regular inscribed polygon of 
double the number of sides, r 



chords in place of one, the polygon APBQC • • • has double the number 
of sides of the polygon ABCD. 

The work on inscribed and circumscribed polygons is essential to 
the understanding of the propositions in connection with the measure¬ 
ment of the circle, as will be shown later. 


A M B 









232 


FUNDAMENTAL THEOREMS 


BOOK V 


Exercises. Inscribed and Circumscribed Polygons 

1. The perimeter of a regular inscribed polygon is less 
than the perimeter of a regular inscribed polygon of double 
the number of sides; and the perimeter of a regular 
circumscribed polygon is greater than that of a regular 
circumscribed polygon of double the number of sides. 

2. Tangents at the midpoints of the arcs 
between adjacent points of contact of the sides 
of a regular circumscribed polygon form a 
regular circumscribed polygon of double the 
number of sides. 

3. The radius drawn to any vertex of a regular polygon 
bisects the angle at the vertex. 

In a square of side s and radius r find the following: 

4. r when s = 8. 5, s when r = 9. 

In an equilateral triangle of side s, radius r, apothem a, 
and area A find the following: 

6. s when r = 4. 8. s when a = Vs. 

7. a when s = Vs. 9. Awhens = V3. 

Find the area of the square inscribed in a circle of radius: 

10. 4in. 11. Gin. 12. 10in. 13. ninches. 

In a regular octagon {§ 90) find the number of degrees in: 

14. The angle at the center. 

15. Each angle of the polygon. 

16. The sum of one angle at the center and one angle 
of the polygon. 

17. The radius of an equilateral triangle is how many 
times the apothem ? what part of the side ? 


w 




§§ 278,279 


SIMILAR REGULAR POLYGONS 


233 


Proposition 4. Similar Regular Polygons 

278. Theorem. Two regular polygons of the same num¬ 
ber of sides are similar. 



Given the regular polygons P and P\ each of n sides. 

Prove that P and P' are similar. 

Proof. Since P and P' are regular, 

then each Z of P = (ti — 2)/n st. Z, 

and each Z of P'= (n — 2)/n st. A. 

P and P' are mutually equiangular. 

Furthermore, a = b = c= d = e, 

and a' = b' = c' = d'=d. 

Then t £. I 1 . 

a b' d d e 

that is, the corresponding sides of P and P' are proportional. 

. *. P and P' are similar. § 205 


Given 

§96 
Ax. 5 

§263 
Ax. 4 


279. Corollary. The areas of two regular polygons of the 
same number of sides are to each other as the squares on any 
two corresponding sides. 

Since the polygons are similar (§ 278), their areas are to each other 
as the squares on any two corresponding sides (§ 251). 

PS 










234 


FUNDAMENTAL THEOREMS 


BOOK V 


Proposition 5. Perimeters of Regular Polygons 

280. Theorem. The perimeters of two regular polygons 
of the same number of sides are to each other as their 
radiiy and also as their apothems. 



Given two regular polygons of n sides, with centers O and O', 
perimeters p and radii r and r' (or OA and O'A'), and apo¬ 


thems a and a' (or OM and O'M') respectively. 

Prove that p:p’=r:r'= a:a'. 

Proof. Draw the radii OR, O'^'. ‘ Post. 1 

Now p:p'=AB:A’B'. §§278,224 

Furthermore, Z.A0B = AA'0'B', §271, Post. 9 

and 0A:0B=1 = 0’A’:0'B’. §134,1 

Hence AOAB and O'A'B’ are similar, § 213 

and AB:A'B' = r:r\ §205 

Also, AAMO and A'M'O^ are similar. § 210 

Hence r:r' = a:a\ § 205 

/. p:p' = r:P=a:a', Ax. 5 


281. Corollary. The areas of two regular polygons of the 
same number of sides are to each other as the squares on the 
radii of the circumscribed circles, and also as the squares on 
the radii of the inscribed circles. 













§§ 280-283 AREA OF A POLYGON* 235 

Proposition 6. Area of a Regular Polygon 

282. Theorem. The area of a regular 'polygon is half 
the product of its apothe'm and its perimeter. 



Given the regular polygon ABCDEF with apothem a, perim¬ 
eter p, and area S. 

Prove that S=i ap. 

Proof. Draw the radii OA, OB, OC, • • • to the successive 
vertices of the polygon, thus dividing the polygon into as 
many congruent A (§ 47) as it has sides. 

The apothem is the common altitude of these A, and the 
area of each A is ja X the base. § 244 

Hence the sum of the areas of all the congruent A is 
^aX the sum of all the bases. Ax. 1 

But the sum of the areas of the A is the area of the poly¬ 
gon, and the sum of the bases is its perimeter. Ax. 10 
,\S=\ap, Ax. 5 

283. Similar Parts. In different circles similar arcs, simi¬ 
lar sectors, and similar segments are such arcs, sectors, and 
segments as correspond to equal angles at the center. 

For example, two arcs of 30° in different circles are similar arcs, 
and the sectors formed by drawing radii to the ends of the arcs are 
similar sectors. 











236 


FUNDAMENTAL THEOREMS 


BOOK V 


Exercises. Regular Polygons 

1. Find the ratio of the perimeters and the ratio of the 
areas of two regular hexagons whose sides are 4 in. and 
8 in. respectively. 

2. Find the ratio of the perimeters and the ratio of 
the areas of two regular octagons whose sides are in the 
ratio 4:2. 

3. Find the ratio of the perimeters of two squares whose 
areas are 484 sq. in. and 121 sq. in. respectively. 

4. Find the ratio of the perimeters and the ratio of the 
areas of two equilateral triangles whose altitudes are 9 in. 
and 36 in. respectively. 

5. The area of one equiangular triangle is 16 times that 
of another. Find the ratio of their altitudes. 

6. The area of the cross section of a steel beam 2 in. 
thick is 24 sq. in. What is the area of the cross section of 
a beam of the same proportions and Ij in. thick ? 

7. Squares are inscribed in two circles of radii 6 in. and 
18 in. respectively. Find the ratio of the areas of the squares, 
and also the ratio of the perimeters. 

8. Squares are inscribed in two circles of radii 6 in. and 
24 in. respectively, and on the sides of these squares equi¬ 
lateral triangles are constructed. What is the ratio of the 
areas of these triangles ? 

9. A square piece of timber is sawed from a round log 
2 ft. in diameter so as to have the cross section of the timber 
the largest possible. What is the area of this cross section ? 
What is the area of the cross section of the largest square 
beam that can be cut from a log of half this diameter ? 

10. Every equiangular polygon inscribed in a circle is 
regular if it has an odd number of sides. 


§§284,285 PUNDAMENTAL CONSTRUCTIONS 


237 


II. Fundamental Constructions 
Proposition 7. Inscribed Square 
284. Problem. Inscribe a square in a given circle. 



Given a O with center 0. 

Required to inscribe a square in the O. 

Construction. Draw any diameter AOC, Post. 1 

At O construct the diameter!)^ J_ to AC. § 104 

Draw AB, BC, CD, and DA, Post. 1 

Then ABCD is the required square. 

Proof. The ACBA, DCB, ADC, BAD are rt. zi• §173 

Since the A at the center are rt. A, Const, 

then the arcs AB, BC, CD, and DA are equal. § 136 

/,AB = BC=CD = DA, §139 

Hence the quadrilateral ABCD is a square. § 15 


285. Corollary. Inscribe regular polygons of 8, 16, 32, 
sides in a given circle. 

By bisecting the successive arcs in the figure of § 284, a regular 
polygon of eight sides may be inscribed in the O. By continuing the 
process regular polygons of how many sides may be inscribed ? 

In general we may say that this corollary allows us to inscribe a 
regular polygon of 21^ sides, where n is any positive integer. 











238 


FUNDAMENTAL CONSTRUCTIONS 


BOOK V 


Proposition 8. Regular Inscribed Hexagon 

286. Problem. Inscribe a regular hexagon in a given 
circle. --- 



Given a O with center 0, 

Required to inscribe a regular hexagon in the O. 

Construction. Draw any radius, as OA. Post. 1 

With A as center and a radius equal to OA, construct an 
arc intersecting the O at B. Post. 4 

Draw AB, Post. 1 

Then AB is a side of a regular hexagon. 

Hence the required hexagon is inscribed by applying AB 
six times as a chord. 

Proof. Draw OB. Post. 1 

Then A OA5 is equiangular. §43 

.*. /LAOB is i of a st. Z, or J of 2 st. A. § 65 

Hence arc AB is i of the O, § 171 

and chord AB is a side of a regular inscribed hexagon. § 272 

287. Corollary. Inscribe an equilateral triangle in a given 
circle. 

Join the alternate vertices of a regular inscribed hexagon. 

288. Corollary. Inscribe regular polygons of 12, 2Jf, Jf8, • 
sides in a given circle. 










§§ 286-290 


INSCRIBED HEXAGON 


239 


289. Extreme and Mean Ratio. If a line segment is divided 
into two segments such that one is the mean proportional 
between the whole line and the other, the line segment is 
said to be divided in extreme and mean ratio. 

The name comes from the fact that one part is a mean and the 
whole line segment and the other part are extremes. 

For example, the line segment of is divided in extreme and mean 
ratio if a segment x is found such that a:x = x:a — x. From this equa¬ 
tion it can be shown that x = 0.618 a. That is, x = 0.6a and a — a; = 0.4 a, 
approximately, so that the division is about 2:3. 

This division of a line segment is often called the Golden 
Sectiony a relatively modern term. At one time, about 1500, 
it was commonly called the Divine Proportion, 

290. Geometric Forms in Art. Since the division of a line 
in the ratio 2:3 is especially pleasing to the eye, the Golden 
Section is often seen in architecture and in the general 
plans of paintings. It is also seen in leaves and flowers. 



Mosaic from Damascus Arabic Pattern 


The use of geometric forms in art is so familiar as to 
require only brief mention. The flgures here shown illus¬ 
trate combinations of regular and semiregular polygons. 

Except for students specializing in mathematics, §§ 289-296 may be 
omitted. They are not generally required in standard courses. 













240 


FUNDAMENTAL CONSTRUCTIONS 


BOOK V 


Proposition 9. Golden Section 

291. Problem. Divide a given line segment in extreme 
and mean ratio. 



Given the line segment AB. 

Required to divide AB in extreme and mean ratio. 


Construction. At B construct a J_ to AB, § 104 

and on it take BO = iAB = BM. §102 

With O as center and BO as radius, construct a O. Post. 4 
Draw AO, meeting the O at D and E. Post. 1 


On AB take AC = AD, and on BA produced take AC'= AE. 
Then C, C' are the required points of division; that is, 
AB:AC = AC:CB, and AB:AC'==AC’:C'B. 


Proof. AE:AB = AB: AD. 

Then, by the laws given in § 198, we have 


222 


AE-AB:AB = 

AB-AD:AD. 
.AE-DE:AB = 


AB-{-AE:AE= 

AD-{-AB:AB. 

.\AB+AC':AC'= 


AB-AC:AC. 

AC:AR=C5:AC. 

.\AB:AC=AC:CB. 


AD+DE:AB. 
.*. C'B:AC'=AC':AB. 
.\AB:AC'=AC':C'B. 













§291 


GOLDEN SECTION 


241 


Exercises. Review 

1. Given an equilateral triangle inscribed in a circle, 
circumscribe an equilateral triangle about the circle. 

2. Given an equilateral triangle inscribed in a circle, 
inscribe a regular hexagon in the circle and circumscribe 
a regular hexagon about the circle. 

3. Divide a line 2 in. long in extreme and mean ratio. 
Measure to the nearest iV in. the lengths of the two seg¬ 
ments of both the internal and the external division and 
compare the results with the ratio given in § 289. 

4. Consider Ex. 3 for a line 2| in. long; a line 3 in. long. 

5. In this illustration from 
a mosaic in an ancient church 
at Constantinople it looks as if 
the broad bands which connect 
the regular hexagons formed 
equilateral triangles. It also 
looks as if the midpoints of the 
sides of these triangles were 
the vertices of other equilateral 
triangles. Investigate these 
two possibilities geometrically. 

6. Find the ratio of the side of an inscribed equilateral 
triangle to the side of a similar circumscribed triangle. 

7. In the internal division of the given line segment in 
§ 291, which part is the mean proportional, the long part 
or the short one? How is it in the case of the external 
division? Write a statement of these two facts. 

8. Find a point within a given triangle such that lines 
from this point to the vertices divide the triangle into three 
equivalent parts. 








242 


FUNDAMENTAL CONSTRUCTIONS 


BOOK V 


Proposition 10. Regular Inscribed Decagon 

292. Problem. Inscribe a regular decagon in a given 
circle, - 



Given a O with center 0. 

Required to inscribe a regular decagon in the O. 
Construction. Draw any radius OA. Post. 1 

Divide OA in extreme and mean ratio; § 291 

that is, so that OA : OP = OP : AP, 

With A as center and OP as radius, construct an arc 


intersecting the O at B. 

Post. 4 

Draw 

AB. 

Post. 1 

Then 

AB is a side of a regular decagon. 


Hence the required regular decagon is inscribed 
plying AB ten times as a chord. 

by ap- 

Proof. Draw PB and OB. 

Post. 1 

Now 

OA:OP=OP:AP, 


and 

AB = OP. 

Const. 


OA:AB=AB:AP. 

Ax. 5 

Moreover, 

ZBAO = ZBAP. 

Iden. 

Then 

A OAB and BAP are similar, 

§213 

and hence 

OA:BA = OB:BP. 

§205 








§§ 292-294 INSCRIBED DECAGON 248 

But OA = OB, § 134,1 

Then BA = BP, §198,2 

and hence BA=BP= OP, Ax. 5 

AAPB=ABAP2ind ZPOB = ZOBP. §42 
But ZAPB = ZPOB-{-ZOBP, §66 

and hence ZBAP=: 2 ZPOB, Ax. 5 

Now ZBAP=ZBAO =ZOBA, Iden., § 42 

and ZPOB=ZAOB, Iden. 

Hence ZBAO =ZOBA = 2 ZAOB, Ax. 5 

and the sum of the A of ZOAB= 5 ZAOB. Ax. 1 
But the sum of these ^ = a st. Z. § 65 

Hence bZAOB=SiSt.Z, Ax. 5 

and 10 ZAOB= 2st A; Ax.3 

whence ZAOB = tV of 2 st. Z. Ax. 4 

Hence arc AB is iV of the O, § 171 


and chord AB is a side of a regular inscribed decagon. § 272 

293. Corollary. Inscribe a regular pentagon in a given 
circle. 

Join the alternate vertices of a regular inscribed decagon. 

From the regular pentagon it is possible to construct the regular 
five-pointed star here shown. 

The Pythagoreans (§ 254), about 525 B.C., are sup¬ 
posed to have been the first to solve the problem of 
constructing a regular pentagon. Because of this fact 
they chose the regular five-pointed star as the badge 
of a brotherhood made up of members of their 
famous school. 

294. Corollary. Inscribe regular polygons of 20,40, 80,--- 
sides in a given circle. 

By bisecting the arcs of the sides of a regular inscribed decagon, 
a regular polygon of how many sides may be inscribed in the O ? By 
continuing the process, regular polygons of how many sides may be 
inscribed in the O? 




244 


FUNDAMENTAL CONSTRUCTIONS 


BOOK V 


Proposition 11. Regular Polygon of 15 Sides 

295. Problem. Inscribe a regular polygon of fifteen 
sides in a given circle. 



Given a O. 

Required to inscribe a regular polygon of 15 sides in the O. 

Construction. From any point A on the O construct a 
chord AC equal to the radius of the O (§ 286), and a chord 
AB equal to a side of a regular inscribed decagon (§ 292). 

In order to obtain a distinct figure only a portion of the G is shown, 
and the detailed construction of the chord AB is assumed from § 292. 


Draw BC, Post. 1 

Then BC is a side of a regular polygon of 15 sides. 
Hence the required polygon is inscribed by applying 
BC fifteen times as chord. 

Proof. Since arc AC is J of the O § 286 

and arc AB is fo of the O, § 292 

then arc BC is tV, or tV of the O. Ax. 2 

Hence chord BC is a side of a regular inscribed polygon 
of 15 sides. § 272 


A polygon of 15 sides is called a pentadecagon, but the term is 
rarely used. 

296. Corollary. Inscribe regular polygons of 30y 60y 120y • * • 
sides in a given circle. 








§§ 295,296 


REGULAR POLYGONS 


245 


Exercises. Regular Polygons 

1. A five-cent piece is placed on the table. How many 
five-cent pieces can be placed around it, each tangent to it 
and tangent to two of the others ? Prove it. 

Circumscribe about a given circle the following regular 
polygons: 

2. Triangle. 4. Hexagon. 6. Pentagon. 

3. Square. 5. Octagon. 7. Decagon. 

Construct an angle of: 

8. 36°. 9. 18°. 10. 9°. 11. 24°. 12. 12°. 

With a side of given length construct: 

13. An equilateral triangle. 16. A regular octagon. 

14. A square. 17. A regular pentagon. 

15. A regular hexagon. 18. A regular decagon. 

19. A regular polygon of fifteen sides. 

20. Prove that the diagonals AC, JSD, CE, DFy EA, FB 
of the regular hexagon ABCDEF form another regular 
hexagon. 

21. Prove that the diagonals AC, BDy CE, DAy EB of the 
regular pentagon ABCDE form another regular pentagon. 

In a regular inscribed polygon in which n is the number 
of sideSy a the apothem, r the radiuSy A an angle^ and C an 
angle at the centery prove the following: 

22. If n = 3, then A = 60°, a=r, and C = 120°. 

23. If 71 = 4, then A = 90°, a = JrV2, and C=90°. 

24. If 77 = 6, then A = 120°, a=^r Vs, and C = 60°. 

25. If 77=10, then A=144°, a=\ 7^'\/l0+2\/5, and C=36°. 



246 


FUNDAMENTAL CONSTRUCTIONS 


BOOK V 


26. Find the perimeter of an equilateral triangle inscribed 
in a circle of radius 3 in. 

27. Find the perimeter of an equilateral triangle circum¬ 
scribed about a circle of radius 2 in. 

28. Find the perimeter of a regular hexagon circum¬ 
scribed about a circle of radius 4 in. 

29. From a circular log with a diameter of 18 in. a builder 
wishes to cut a column with its cross section as large a 
regular octagon as possible. Find the length of a side of 
the cross section. 

30. In the figure here shown ABCD is a D ^ 

square. Arc POQ is constructed as part p 
of a circle with center A and radius AO, 
and the other arcs are constructed in a 
similar manner. Prove that the octagon ^ ^ 

seen in the figure is regular. ^ 

31. The area of a regular inscribed hexagon is what part 
of the area of a regular hexagon circumscribed about the 
same circle? 

32. Construct a regular pentagon, given one of the 
diagonals. 

33. In a given equilateral triangle inscribe three equal 
circles, tangent each to the other two and to two sides of 
the triangle. 

34. The points A, J5, C, A * * * are consecutive vertices of 
a regular inscribed octagon, and A, B\ C\ D', • • • are con¬ 
secutive vertices of a regular polygon of twelve sides in¬ 
scribed in the same circle. Find the angle formed by each 
pair of the following lines, produced if necessary: 

(1) AB and AB'. (3) AB and AC. (5) AB' and AD. 

(2) AB and AC'. (4) AB and AD. (6) B'C' and AC. 





§297 


CIRCLE MEASUREMENT 


247 


III. Circle Measurement 

297. Plan of Measurement. For practical purposes we can 
find the circumference of a circle very easily. If we wind 
a piece of paper about a cylinder, prick through the paper 
with a needle where the paper overlaps, and then fiatten 
the paper out on a table, we can measure with a fair degree 
of accuracy the distance between the two points thus made. 
Evidently, however, this is not as accurate as the measure¬ 
ment of a straight line by means of a pair of dividers or 
compasses, because paper tends to stretch or to contract. 

For scientific purposes we therefore resort to mathe¬ 
matics. One reason for showing how to inscribe and cir¬ 
cumscribe regular polygons, and then to double the number 
of sides, is to construct polygons that approach nearer and 
nearer to the circle. Since we can measure these polygons, 
both as to perimeter and as to area, we can thus approxi¬ 
mate the circumference, and can also approximate the area 
which the circle incloses. We may carry this approximation 
to any degree of accuracy that we wish. 

For example, if we find the perimeter of an inscribed 
square, then find the perimeter of an inscribed regular 
octagon, and continue this process for polygons of 16, 32, 
64, • • • sides, we can find a perimeter which approaches as 
near the circumference as we choose, and similarly for the 
area inclosed by the circle. 

In this way we can find the approximate ratio of the circumference 
of a circle to its diameter. The student who takes up the calculus in 
college will there find a simpler method of solving this problem. 

We shall, therefore, first consider the problem of finding 
the perimeter of a regular polygon of double the number 
of sides of a given regular polygon; or, what is more simple, 
of finding one side of such a polygon. 


248 


CIRCLE MEASUREMENT 


BOOK V 


Proposition 12. Doubling the Sides 

298. Problem. Given the side and the radius of a 
regular inscribed polygon^ find the side of a regular 
inscribed polygon of double the number of sides. 



Given 5 (or AR), the side, and r, the radius, of a regular 
polygon inscribed in the G with center 0 , 


Required to find a side of a regular inscribed polygon of 
double the number of sides. 


Solution. 

Construct PQ, the J_ bisector of s. 

§§ 102, 104 

Draw 

AP and AQ, 

Post. 1 

Then 

PQ is a diameter and bisects arc AB, 

§143 


AP is the required side. 

§277 

Since AM=is, ONf=r^—\s^; 

§253 

whence 

OM = V r^— 1 

Ax. 6 


:.PM=r-OM=r- Vr"- \ 

Ax. 5 

Further, 

A AMP and QAP are similar, 

§210 

and hence 

PM:AP=AP:PQ, 

§205 

Then 

AP‘=PQ-PM-, 

§ 198,1 

whence 

AP^=2r{r — ^7^—\s^). 

Ax. 5 

Hence 

AP=a/2 r (r — Vr ^—5 s^). 

Ax. 6 

or 

AP=V^(2r — V4r^— s^). 


















§§ 298-301 


LIMITS 


249 


299. Constant and Variable. If we inscribe a regular poly¬ 
gon in a given circle, and then continue to double the 
number of sides of this polygon, the perimeter continues 
to vary in size, approaching nearer and nearer the circle, 
which remains constantly the same in size. A quantity 
considered as having a fixed value throughout a given dis¬ 
cussion is called a constant, and a quantity considered as 
having different successive values is called a variable. 

In the above case, the perimeter of the polygon, as we increase the 
number of sides, is a variable, but the circle is a constant. 

300. Limit. When a variable so approaches a constant that 
the difference between the two may become and remain 
less than any assigned positive quantity, however small, 
the constant is called the limit of the variable. 

Sometimes variables can reach their limits and sometimes they 
cannot. For example, a chord may increase in length up to a certain 
limit, the diameter, and it can reach this limit and still be a chord; 
it may decrease, approaching the limit 0, but it cannot reach this limit 
and still be a chord as we define it in elementary work. 

If p is the perimeter of a regular inscribed or of a regular circum¬ 
scribed polygon and c is the circle, we say that ''p tends to c,’' or 

approaches c as its limit,” indicating this by the symbol p— 

301. Principles of Limits. From the above definition we 
may assume as postulates the following principles: 

1. If a variable x approaches a finite limit Z, and if c is a 

constant, then cx approaches the limit cl, and - approaches 

I ^ 

the limit - • 

c X I 

That is, if x—^l, then cx — ycl and 

2. If, while approaching their respective limits, two vari¬ 
ables are always equal, their limits are equal. 

For if the limits were unequal, the two variables would be unequal 
when they were very near their limits. 

PS 


250 


CIRCLE MEASUREMENT 


BOOK V 


302. Area of a Circle. The area of the space inclosed by 
a circle is called the area of the circle. 

With the modern definition of a circle as a line, the expression 
'' area of a circle ” has no meaning unless it is specifically defined. 
We therefore define it as a brief form of the longer expression area 
inclosed by a circle.’^ 

303. Limits related to the Circle. From what has been said 
concerning the circle and the regular inscribed polygon we 
may assume as true the following statements: 

1. The circumference of a circle is the limit of the perim¬ 
eter of a regular inscribed or of a regular circumscribed 
polygon as the number of sides is indefinitely increased, 

2. The area of a circle is the limit of the area of a regular 
inscribed or of a regular circumscribed polygon as the number 
of sides is indefinitely increased. 

3. If the number of sides of a regular inscribed polygon is 
indefinitely increased, the apothem of the polygon approaches 
the radius of the circle as its limit. 

In this figure, if n is the number of sides of the 
polygon, then a — >ON as n — ^co; that is, a ap¬ 
proaches ON as its limit as the number of sides 
increases without limit. We are not justified in 
saying that the expression n —>-oo means that n 
approaches infinity as a limit, because the word 
'' infinity ” means without limit. We may, however, say that n tends 
to infinity ” or that '' n approaches infinity.” 

In higher mathematics the statements given above are proved with 
the same care with which we prove a proposition in the geometry of 
rectilinear figures, but in an elementary treatment of measurement it 
is impossible to give satisfactory proofs; indeed, the truth of the 
statements would be no more evident if the proofs were given. By 
informal discussion their truth is as apparent as that of any postulate. 

In the case of a regular circumscribed polygon the apothem is always 
the same as the radius of the circle, and hence, with this fact under¬ 
stood, we may say that all three assumptions apply to either inscribed 
or circumscribed regular polygons. 




§§ 302-307 


CIRCUMFERENCES 


251 


Proposition 13. Ratio of Circumferences 

304. Theorem. Two circumferences have the same ratio 
as the radii. 



Given the © O and O' with circumferences C and C' and 
radii r and r' respectively. 

Prove that C:C'=r:r', 

Proof. Let he the perimeters of two similar regular 


inscribed polygons. § 269 

Then p:p'=r:r', §280 

/,pr' = p'r, ^ §198,1 

Let the number of sides be increased uniformly. 

Then p-^C, and C\ § 303,1 

and hence Cr'=C'r, §301 

. ,\C:C'=r:r\ §198,3 


305. Corollary. The ratio of any circle to its diameter is 
constant. 

Since C:C' = 2r:2r\ then C:2r=C''.2r'. 

306. Symbol tt. The constant ratio of a circle to its diam¬ 
eter is represented by the Greek letter tt (pi). 

307. Corollary. In any circle^ C—2irr = n-d. 

c c 

By definition (§ 306), tt = — = - ; whence C = 2 7rr and C = Trd. 

2r d 







252 


CIRCLE MEASUREMENT 


BOOK V 


Proposition 14. Area of a Circle 

308. Theorem. The area of a circle is half the product 
of the radius and the circumference. 



Given the O 0 with radius r, circumference C, and area A, 
Prove that A = ^ rC, 

Proof. Circumscribe about the O a regular polygon of 
n sides, and let p be its perimeter and A' its area. § 270 


Then 

Let 

Since 

and 

then 

Also, 

But, always. 


A’-- 


irp. 


n be increased indefinitely. 
p^C 

r is constant. 


(282 


i 303,1 


\rp- 


-irC. 


§ 301,1 
§ 303, 2 
§282 
§ 301, 2 


A'-^A. 

A'=^rp. 

A = lrC. 

309. Corollary. The area of a circle is tt times the square 
on the radius. 

For A = ^ rC = I r X 2 7rr = Trr^. 

310. Corollary. The areas of two circles are to each other 
as the squares on the radii. 







§§ 308-310 


CIRCUMFERENCE AND AREA 


253 


Exercises. Circumference and Area 

1. The area of a sector is half the product of the radius 
and the arc, 

2. If the circumference of one circle is twice that of 
another, the square on the radius of the first is how many 
times the square on the radius of the second ? 

3. If the circumference of one circle is four times that 
of another, an equilateral triangle constructed on the 
diameter of the first as side has how many times the area 
of an equilateral triangle constructed on the diameter of 
the second as side ? 

4. A water pipe with a diameter of 3 in. has a circum¬ 
ference of 9.425 in. Find the circumference of a water pipe 
which has a diameter of 4 in. 

5. A wheel with a circumference of 8 ft. has a diameter, 
expressed to the nearest 0.01 ft., of 2.55 ft. Find the cir¬ 
cumference of a wheel with a diameter of 3.175 ft. 

6. A regular hexagon is 4 in. on a side. Find both its 
apothem and its area to the nearest 0.01. 

7. If the radius of one circle is four times that of 
another, and if the area of the smaller circle is 31.4 sq. in., 
what is the area of the larger circle ? 

8. If the radius of one circle is five times that of 
another, and if the area of the smaller circle is 9.6 sq. in., 
what is the area of the larger circle ? 

9. The circumferences of two cylindric steel shafts are 
7 in. and 3j in. respectively. The area of the cross section 
of the first shaft is how m^ny times that of the second ? 

10. If the arc of a sector of a circle 3j in. in diameter is 
2 in. long, what is the area of the sector ? 

Use Ex. 1, above, in finding the required area. 



254 


CIRCLE MEASUREMENT 


BOOK V 


Proposition 15. The Value of tt 

311. Problem. Find the approximate value of the ratio 
of the circumference of a circle to its diameter. 

Given a O with circumference C and diameter d. 

Required to find the approximate value of tt. 

Solution. Let Sq be the length of a side of a regular 
inscribed polygon of 6 sides, of 12 sides, and so on. 

The student need not perform the computations or recall the follow¬ 
ing steps, but he should understand the general nature of the work. 

Then Sj 2 = '\/r{2r — y/4:F — si). §298 

But, when r = 1, Se = 1* § 286 

Hence, using the successive values of s, we have 


Form of Computation 

Length of Side 

Perimeter 

s.,=V2-V4-r 

0.51763809 

6.21165708 

= V 2 - \/4 - 0.51763809" 

0.26105238 

6.26525722 

S 48 = V 2 - V4 - 0.26105238" 

0.13080626 

6.27870041 

Sge = ■\/2 - V4 - 0.13080626" 

0.06543817 

6.28206396 

s.*,=V 2 - V4 - 0.06543817" 

0.03272346 

6.28290510 

8394 =V 2 - \/4 - 0.03272346" 

0.01636228 

6.28311544 

s,e»=V 2 - V4 - 0.01636228" 

0.00818121 

6.28316941 

Since C = 

27rr, 

§307 


when r = 1, tt = C. 

But, when n = 768, C = 6.28317, approximately, 

and hence tt = 3.14159, approximately. 

For thousands of years the world tried to find the value of the 
incommensurable number tt. The ancients generally considered the 
value as 3 or as 3b We generally use the following values: tt = 3.1416, 
or 3b and I/tt = 0.31831. 


















§311 


EXERCISES 


255 


Exercises. Circle Measurement 

Find the circumferences of circles with radii as follows: 

1. 2 in. 3. 3.2 in. 5. 6iin. 7. 3 ft. 6 in. 

2. 3 in. 4. 4.3 in. 6. 7f in. 8. 4 ft. 2 in. 

In all the work on this page use the value 3.1416 for tt. 

Find the circumferences of circles with diameters as 
follows: 

9. 4 in. 11. 6.2 in. 13. 3| ft. 15. 30 cm. 

10. 22 in. 12. 8.3 in. 14. 2|in. 16. 42 mm. 

Find the radii of circles with circumferences as follows: 

17. 3-77. 19. 15.708 in. 21. 18.8496. 23. 345.576. 

18. 47r. ' 20. 21.9912 ft. 22. 125.664. 24. 3487.176. 

Find the diameters of circles with circumferences as 
follows: 

25. 87 r. 27. 27rr. 29. 188.496in. 31. 3361.512in. 

26. tt". 28. 37ra". 30. 219.912 in. 32. 3173.016 in. 

Find the areas of circles with radii as follows: 

33. 2x. 35. 16 ft. 37. 4jin. 39. 3 ft. 4in. 

34. 37r. 36. 5.8 ft. 38. 3|in. 40. 5 ft. 8 in. 

Find the areas of circles with diameters as follows: 

41. 10 ah. 43. 3.5 ft. 45. 2f yd. 47. 2 ft. 4 in. 

42. 12 7^^ 44. 4.3 in. 46. 3i yd. 48. 3 ft. 6 in. 

Find the areas of circles with circumferences as follows: 
49. 3 7 r. 50. tt/c. 51. 18.8496 in. 52. 333.0096 in. 

Fhid the radii of circles with areas as follows: 

53. 'tto^, 54. TT. 55. 12.5664. 56. 78.54. 


256 


CIRCLE MEASUREMENT 


BOOK V 


Exercises. Applications 

1. The diameter of a bicycle wheel is 28 in. How many 
revolutions does the wheel make in 8 mi.? 

2. Find the diameter of an automobile wheel which 
makes r revolutions in half a mile. 

3. A circular pond 200 yd. in diameter is surrounded by 
a walk 8 ft. wide. Find the area of the walk. 

4. The span (chord) of a bridge in the form of a circular 
arc is 60 ft., and the highest point of the arch is 7 ft. 6 in. 
above the piers. Find the radius of the arch. 

5. Two branch drain pipes lead into a main drain pipe. 
It is necessary that the cross-section area of the main pipe 
shall equal the sum of the cross-section areas, of the two 
branch pipes, which are respectively 6 in, 
and 8 in. in diameter. Find the diameter of 
the main pipe. 

6. The top part of the kite here shown is 
a semicircle and the lower part is a triangle. 

Find the area of the kite. 

7. In making a drawing for an arch it is 
necessary to mark off on a circle drawn with 
a radius of 10|- in. an arc that shall be 11 in. long. This is 
best done by finding the angle at the center. How many 
degrees are there in this angle? 

8. In the iron washer here shown, the 
diameter of the hole is 2f in. and the width 
of the metal ring is fin. Find the area of 
one face of the washer. 

9. Find the area of a fan which opens out into a sector 
of 120° with a radius of 10 in. 

10. Consider Ex. 9 for a radius of 5 in. 










§311 


GENERAL REVIEW 


257 


IV. General Review 
Exercises. Review 

Write a classification of the different kinds of: 

1. Lines. 3. Triangles. 5. Polygons. 

2. Angles. 4. Quadrilaterals. 6. Parallelograms. 

State the conditions under which: 

7. Two triangles are congruent; are equal in area; are 
similar. 

8. Two straight lines are parallel. 

9. Two parallelograms are equal in area. 

10. Two polygons are similar. 

Complete the following statements in general terms: 

11. In a right triangle the square on the • • 

12. If two parallel lines are cut by a transversal, • • •. 

13. An angle formed by two secants drawn to a circle 
is measured by • • •. 

14. The perimeters of two similar polygons are to each 
other as • • •, and their areas are to each other as • • •. 

15. Equal chords of the same circle or of equal circles • • •. 

16. Two central angles of the same circle or of equal 
circles have • • •. 

17. If two secants intersect within, on, or outside a circle, 
the product of • • *. 

18. The sum of the interior angles of • • •. 

19. The area of a polygon is • • •. 

20. One formula for the • • • of a circle is I ird^. 

21. One formula for a • • • is 7rr. 


258 


GENERAL REVIEW 


BOOK V 


Exercises. Loci 

1. Find the locus of the center of the circle inscribed 
in a triangle which has a given base and a given angle at 
the vertex. 

2. Given a line segment, find the locus of the end of a 
tangent to a given circle such that the length of the 
tangent is equal-to the length of the given segment. 

3. Find the locus of a point from which tangents drawn 
to a given circle form a given angle. 

4. Find the locus of the intersection of the perpendic¬ 
ulars from the three vertices to the opposite sides of a tri¬ 
angle which has a given base and a given angle at the vertex. 

5. Find the locus of the midpoint of a line segment 
drawn from a given point to a given line. 

6. Find the locus of the vertex of a triangle which has 
a given base and a given altitude. 

7. Find the locus of a point such that the sum of its 
distances from two given parallel lines is constant. 

8. Find the locus of a point such that the difference 
between its distances from two given parallel lines is 
constant. 

9. Find the locus of a point such that the sum of its 
distances from two given intersecting lines is constant. 

10. Find the locus of a point such that the difference 
between its distances from two given intersecting lines is 
constant. 

11. Find the locus of a point such that its distances from 
two given points are in the ratio 3:4. 

12. Find the locus of a point such that its distances from 
two given parallel lines are in the ratio m : n. 


§311 


EXERCISES 


259 


Exercises. Constructions 

1. In a given circle inscribe a regular polygon similar 
to a given regular polygon. 

2. Divide the area of a given circle into two equivalent 
parts by a circle which has the same center as the given 
circle. 

3. Construct a circle with its circumference equal to the 
sum of the circumferences of two circles of given radii. 

4. Construct a circle with its circumference equal to 
the difference between two circumferences of given radii. 

5. Construct a circle with its area equal to the sum of 
the areas of two circles of given radii. 

6. Construct a circle such that its area is three times 

the area of a given circle. • 

7. Construct a circle such that the ratio of its area to 
that of a given circle \s>m:n. 

8. In a given square inscribe four equal circles such 
that each circle is tangent to two of the others and to two 
sides of the square. 

9. In a given square inscribe four equal circles such 
that each circle is tangent to two of the others and to one 
side and only one side of the square. 

10. Construct a common secant to two given circles, 
which are exterior to each other, such that the intercepted 
chords shall have the given lengths a and 6. 

11. Through a point of intersection of two given inter¬ 
secting circles construct a common secant of a given length. 

12. Construct a tangent to a given circle such that the 
segment intercepted between the point of contact and a 
given line has a given length. 


260 


GENERAL REVIEW 


BOOK V 


Exercises. Formulas 

If r is .the radius of a circle, s one side of a regular 
inscribed polygon, and n the number of sides, prove the 
following, and find s to the nearest 0.01 when r==l: 

1. If n = 3, s = r Vs. 4. If 7 z = 5, g = |r\ /l0 —2 V5. 

2. If = 4, s = r V 2 . 5. If n = 8, s = r ^2 — V2. 

3. li n = Q, s = r. 6. If 7t = 10, § = 1 r(V5 —l). 

7. If a regular pentagon of side s is inscribed in a circle 

of radius r, find the apothem. 

8. If a regular polygon of side s and apothem a is in¬ 
scribed in a circle of radius r, prove that 

a = V4r^— 

9. A regular polygon of side s is inscribed in a circle of 
radius r. If a side of the similar circumscribed regular 
polygon is s', prove that 

2gr ^ 

V4 f^—s^ 

10. Three equal circles are constructed, each tangent to 
the other two. If the common radius is r, find the area 
inclosed by the arcs between the points of tangency. 

11. Given p and P, the perimeters of regular polygons of 
n sides respectively inscribed in and circumscribed about 
a given circle of radius r, find p' and P', the perimeters of 
regular polygons of 2 n sides respectively inscribed in and 
circumscribed about the given circle. 

12. A circular plot of land a feet in diameter is surrounded 
by a walk b feet wide. Find the area of the circular plot 
and the area of the walk. 

13. In Ex. 12 find the circumference at the outer edge 
of the walk. 







§311 


EXERCISES 


261 


Exercises. Review 

1. The segment which joins the midpoints of the diag¬ 
onals of a trapezoid is equal to half the difference between 
the bases. 

2. If from any point on a circle a chord and a tangent 
are drawn, the perpendiculars drawn to them from the 
midpoint of the minor arc are equal. 

3. Consider Ex. 2 with respect to the midpoint of the 
major arc. 

4. If two equal chords are produced to meet outside a 
circle, the secants thus formed are equal. 

5. If squares are constructed outwardly on the six sides 
of a regular hexagon, the exterior vertices of these squares 
are the vertices of a regular polygon of twelve sides. 

6. The sum of the perpendiculars drawn to any tangent 
to a circle from the ends of a diameter is equal to the 
diameter. 

7. No oblique parallelogram can be inscribed in a circle. 

An oblique parallelogram has oblique angles (§ 16). 

8. Two points C and D are taken on a semicircle of 
diameter AB. If AD and BC meet in E, and AC and BD meet 
in F, then EF is _L to AB. 

9. If the tangents from a given point P to three given 
circles which do not intersect are all equal, the circle drawn 
with center P and passing through the points of contact of 
these tangents cuts the given circles at right angles. 

Two circles are said to intersect at right angles if their tangents at 
a point of intersection are perpendicular to each other. 

10. State and prove the converse of the proposition that 
the square on the hypotenuse of a right triangle is equiv¬ 
alent to the sum of the squares on the other two sides. 


262 


GENERAL REVIEW 


BOOK V 


Exercises. Applications 

1. On a railway curve which is the arc of a circle two 
points P and Q are taken and the chord PQ is found to be 
400 ft. The distance from the midpoint of the arc to the 
midpoint of the chord is 28 ft. Find the radius of the circle. 

2. Two rectangular city lots have the same depth, the 
frontage of the first is twice that of the second, and their 
combined frontage is equal to their common depth. Find 
the ratio of their areas and the ratio of their perimeters. 

3. A ladder 50 ft. long reaches a window 40 ft. from the 
ground on one side of a street, and when tipped backward 
to rest against the building on the opposite side it reaches 
a window 30 ft. from the ground. How wide is the street ? 

4. Two wheat bins of the same height are respectively 
8 ft. and 10 ft. square on the bottom. Find the dimensions 
of the square bottom of a third bin which has the same 
height as each of the other two and the same volume as 
the other two combined. 

5. Two forces of 180 lb. and 240 lb. make an angle of 90° 
with each other. Compute the resultant. 

The resultant is represented graphically by the diagonal of a rec¬ 
tangle of sides 180 and 240. See Ex. 6, p. 106. 

6. In laying out a park it is desired to plant eight trees 
equidistant from one another and each 200 ft. from a foun¬ 
tain. Construct a figure with all construction lines to show 
how the trees should be placed. 

7. A water main is to be laid to two branch pipes which 
have diameters of 12 in. and 18 in. respectively. The diam¬ 
eter of the main must be such that the area of its cross 
section is equal to the sum of the cross-section areas of the 
branches. Find the diameter of the main to the nearest I in. 


§311 


EXERCISES 


263 


Exercises. Review 

1. If the three points of tangency of a circle inscribed 
in a triangle are joined, the angles of the resulting triangle 
are all acute. 

2. If two consecutive angles of a quadrilateral are right 
angles, the bisectors of the other two angles of the quadri¬ 
lateral form a right angle. 

3. The two line segments which join the midpoints of 
the opposite sides of a quadrilateral bisect each other. 

4. If two triangles have equal bases and equal angles 
at the vertex, the areas of the circumscribed circles are 
equal. 

5. If two circles are concentric, the segments inter¬ 
cepted between them on any line are equal. 

6. If any two consecutive sides of an inscribed hexa¬ 
gon are respectively parallel to their opposite sides, the 
remaining two sides are parallel. 

7. The lines which bisect any angle of an inscribed 
quadrilateral and the exterior angle at the opposite vertex 
intersect on the circle. 

8. In order that a parallelogram can be circumscribed 
about a circle, the parallelogram must have equal sides. 

9. The area of a triangle is half the product of its 
perimeter and the radius of the inscribed circle. 

10. The perimeter of a triangle is to any side as the 
altitude from the opposite vertex of the triangle is to the 
radius of the inscribed circle. 

11. If two equivalent triangles have the same base and 
lie on the same side of this base, any line which cuts the 
triangles and is parallel to the base cuts off equal areas 
from the triangles. 


264 


GENERAL REVIEW 


BOOK V 


12. In the triangle whose sides are 10, 36, and 40 com¬ 
pute the length of the projection of the longest side upon 
the shortest side. 

13. Within a rhombus ABCD, in which A and C are oppo¬ 
site vertices, the point P is chosen so that PB=PD. Prove 
that A, P, and C are in the same straight line, and that 

ap-pc=ab‘-pb\ 

14. An isosceles A ABC is inscribed in a circle, and from 
the vertex A a chord AD is drawn to cut the base BC in the 
point E. Prove that AB^—AE^=BE • CE. 

15. In an acute A ABC the altitudes BD and CE inter¬ 
sect in the point O. Prove that OB:OC = OE : OD. 

16. From an external point P two secants are drawn, 
one cutting the circle at the points A and 5, and the other 
at the points C and A so that PA = 5 in., AB = 35 in., and 
PC = CD. Find the length of PD. 

17. The sum of the perpendiculars drawn to the sides of a 
regular polygon from any point within the polygon is equal 
to the product of the apothem and the number of sides. 

18. Find the perimeter and the area of a regular octagon 
inscribed in a circle with a diameter of 32 in. 

19. On the sides of a square ABCD of side a, the points 
P, Q, P, S are taken such that AP — BQ==CR = DS = ^a. 
Prove that PQRS is a square and then find its area. 

20. Each side of a triangle is 2 a inches, and about each 
vertex as a center a circle is constructed with a radius 
of a inches. Find the area bounded by the three arcs 
which lie outside the triangle, and the area bounded by 
the three arcs which lie inside the triangle. 

21. Every equilateral polygon circumscribed about a cir¬ 
cle is regular if it has an odd number of sides. 


§311 


APPLICATIONS 


265 


Exercises. Miscellaneous Applications 

1. Extend your arm toward a distant object, and, closing 
your left eye, sight across a finger tip with your right 
eye. Now keep the finger in the same position and sight 
with your left eye. The finger then seems to point to an 
object some distance to the right of the one at which you 
were pointing. If you can estimate the distance between 
these two objects, which can often be done with a fair 
degree of accuracy when there are houses between them, 
then your distance from the objects is approximately ten 
times the estimated distance between them. Draw a plan 
which shows that the lines of sight are sides of triangles, 
and explain the geometric principle involved. 

2. The distance across a stream can be found by the 
principle involved in any one of these 
three diagrams. Explain the method 
in each case and state the geometric 
principles involved. 

3. An instrument like the one here shown is used 
in measuring heights. The base is graduated in equal 
divisions, say 50, and the upright arm 
is similarly divided. At each end of the 
hinged bar is a sight. If an observer lying 
50 ft. from a tree sights at the top, and 
finds that the hinged bar cuts the upright 
arm at 27, he knows that the tree is 
27 ft. high. Explain the geometric principle involved. 

4. If three streets intersect as here shown, find the area 
of the shaded triangle. 

Use the formula in Ex. 1, page 194. 

5. Can the triangle of Ex. 4 be a 
right triangle ? Prove your answer. 




\ ^ 





PS 














266 


GENERAL REVIEW 


BOOK V 


6. If a dangerous shoal lies near a headland, the verti¬ 
cal danger angle is the angle {AHAX) between the level of 
the water and the line of sight to the h 

headland H from any point, as A, on a _ 

circle of sufficient radius to inclose the pg 
dangerous area. In order to avoid the 

shoal, ships coming near the headland ^ ^ 

should be careful to keep far enough away, say at S, so 
that the AHSX is less than the known danger angle. 
Explain the geometric principle involved. 

7. On his voyage to Egypt, Napoleon is said to have sug¬ 

gested to his staff the problem of dividing a circle into four 
equal parts by the use of circles alone. It is also said that 
the problem was solved by using the figure here shown. 
How was it done? a 

Prove that the area of OR is one fourth that of 
(DA. Then prove that the sum of the four areas 
marked D is equal to the sum of the four areas 
marked C. Then prove that one of the D’s, the 
white part of one of the B’s, and one of the C-s 
together make one fourth of OA. 

8. In locating the site for a union-school building for 
three villages A, B, and C, it is desired to place the school 
so that it shall be equidistant from the three villages. If A 
is 41 mi. from B and 6 mi. from C, and B is 5| mi. from C, 
draw a map to the scale of 1 in. = 1 mi. and show the 
location for the school. 

While in practice the established roads between the villages would 
have to be considered, it may be assumed here that all distances are 
measured in a straight line. 

9. By measuring the map in Ex. 8, find how far it will 
be from each village to the school, and check your answer 
by the formula given in Ex. 5, page 198. 



§311 


APPLICATIONS 


267 



10. If a carpenter's square is placed on top of an upright 
stick, as here shown, and an observer ^ 
sights along the arms to a distant point 
B and to a point A near the stick, then 

if AD and DC are measured, the length ___ 

of DB can be found. Show how this ^ ^ ^ 

can be done, explaining the geometric principle involved. 

Roman surveyors knew this method two thousand years ago. 

11. Surveyors sometimes lay off a right angle in a field 

by setting two stakes P and 5 on a \P 

line 3 ft. apart. They then hold the 

end of a tape at B and the 9-foot AA^:^^ - - 

mark at P, stretch the tape taut a 

toward A, and set a stake at A on the 5-foot mark. Prove 
that ZP is a right angle. 

12. The captain of a ship which is sailing on the course 
ABX observes a lighthouse L when the ship is at A, and 
measures ZA. He then observes the lighthouse until the 
angle at B is just twice that at A. 

He determines the distance AP from 
his log, an instrument which tells 
how far a ship has gone. He then 
knows that PL, the distance from 
the lighthouse, is the same as AP, the distance sailed. 
State the geometric principle involved in this method, 
which is known as ''doubling the angle on the bow." 

13. The rectangular frame here shown has a plumb line 








/ 


P 


->x 


ML 


from 


M 


the upper strip of wood. Prove that 

/ 

% 

when the point of the plumb bob is 



at the midpoint of AP, the base of the 

V 

1 y 












268 


GENERAL REVIEW 


BOOK V 


14. A draftsman’s triangle is placed over two nails 
driven into a board at A and B. If a pencil point is placed 
at P, it will mark an arc of a circle 
as the triangle is moved about so that 
the arms of ZP always touch the two 
nails. State the geometric principle 
involved. 

15. In Ex. 14, if P is taken as the vertex of the inside 
right angle of the triangle and its arms always touch A 
and By what kind of arc is formed upon AB as chord ? 

16. In laying out the tracks for a street railway, which 
is to turn a right-angled corner as shown in this plan, the 
curve is to be tangent to both the 
vertical tracks and to the horizontal 
tracks. The curve is also to be as 
large as possible without running 
the inside track beyond the corner C. 

Show how to find the center of curva¬ 
ture; that is, the center O from which 
the arcs for the curve are drawn. 

17. If an engineer has to extend a curve which he knows 

is an arc of a circle, but which is too large to be drawn 
with a tapeline, or which cannot be ^ 

easily reached from the center, the 
following method is sometimes used : 

Take P as the midpoint of the known \ 

part APB of the curve. Then stretch 
the tape from A to B and construct PM ± to AB, Then 
swing the length AM about P, and the length PM about P, 
until they meet at L, and stretch the length AB along PL 
to Q, thus fixing the point Q. The point C is fixed in the 
same way, and so on for as many points as are necessary. 
Explain the geometric principle involved. 













§311 


APPLICATIONS 


269 


18. In shops where two pulleys are driven by belting, 
we have a case of two tangents to two given circles. If 
the belt runs straight between the pulleys, we have the 
case of two exterior tangents. If the belt is crossed so that 



the pulleys turn in opposite directions, we have the case of 
two interior tangents. In case the belt is liable to change 
its length, on account of stretching or variation in heat or 
moisture, a third pulley C is often used. We then have the 
case of tangents to three pairs of circles. Construct the 
figure for each of the three cases. 

19. This figure shows how a circular 
driveway was laid out from a gate G 
to a porch P so as to avoid a group of 
rocks R. Explain how the plan was 
constructed and state the geometric 
principles involved. 



20. In making the plans for a park a landscape architect 
wished to connect two parallel roads R and R' by the curve 
here shown, which consists of two arcs and is known 
as a reversed curve. From the 
figure explain how the architect 
proceeded to construct the plan, 
and state the geometric prin¬ 
ciples involved at each step. 

The architect located the center 
line of the curve, the dot-and-dash line, before drawing the lines which 
represent the sides of the road. Considering the center line, notice 
that each arc is tangent to a road and that the arcs are tangent to 
each other. 















270 


GENERAL REVIEW 


BOOK V 


21. A draftsman who wished to draw one long line 
perpendicular to another used his T-square 
in the two positions shown in the figure, 
instead of using a triangle. State the 
geometric principles involved in drawing 
the lines in this way. 

22. This instrument is used for drawing a line parallel 
to the edge of a board. Block B is fastened to the end of 
bar E and has a sharp marking 
point on its underside. Block A 
can be clamped in any position 
on bar E by the set screw C. If 
block A is moved along one edge of 
the board, will the point on B trace 
a line parallel to the edge ? Why ? 

23. The gauge in Ex. 22 is also used for dividing a board 
into two equal parts. The equal brass arms AD and BD are 
pivoted at i) by a marking point, and are also pivoted at A 
and B. Blocks A and B are set to the width of the board 
to be divided, and then block A is moved along one edge 
of the board while point D traces the dividing line. State 
the geometric principle involved. 




AB 


24. In turning a piston ring for an engine a larger ring 
is made than is needed in the cylinder. Usually the outside 
diameter of the ring is made 1.5% longer than 
the diameter of the cylinder. The piece AB is 
then cut out, the ring is drawn together at P, 
as shown by the dotted lines, and is fitted in 
place. If the diameter of the cylinder is 4 in., 
what diameter should be used in turning the 
ring and what length should be cut off (AB) to make the 
ring fit the cylinder ? 



















§311 


APPLICATIONS 


271 


Cr-. 



25. In surveying it is often necessary to run a straight 
line beyond an obstacle through which it is impossible to 
sight and over which it is impossible to pass. One of the 
methods, which is illustrated by the 
adjoining figure, is as follows: Sup¬ 
pose that the surveyor desires to 
run the line AB beyond the house H ; 
he first runs a line BC at right angles 
to AB\ at C he runs a line CD at right angles to BC; at D 
he runs a line DX at right angles to CD; on DX he lays off 
DE= CB, and at E he runs a line EF at right angles to DE. 
Prove that EF is part of the straight line AB prolonged. 

26. An 8-inch pipe can carry how many times as much 
water as a 1-inch pipe ? as a 2-inch pipe ? as a 4-inch pipe ? 

In an 8-inch pipe the internal diameter is 8 in. 


27. The diameter of the safety valve of a boiler is 2| in. 
Find the total pressure of the steam upon the face of the 
valve when the steam gauge indicates that the pressure is 
140 lb. per square inch. 

28. The drive wheel of a locomotive is 6 ft. in diameter 
and makes 1722 revolutions while the locomotive is going 
6 mi. Find the distance lost through the slipping of the 
wheel on the track. 

29. The ''dip of the horizonis the Z.TEH in this figure. 
It is the angle formed at the eye E of an observer by the 
line EH which is J_ to OE, the earth's radius 

. produced, and ET, the tangent from E to the 
sea horizon. Prove that the dip of the horizon 
is equal to the Z O at the center of the earth. 

The proportions of such a figure are necessarily ex¬ 
aggerated in drawing. Those who have studied physics 
will also observe that in practice the question of the bending of the 
light rays must be considered. 






272 


GENERAL REVIEW 


BOOK V 


Exercises. College Entrance Examinations 

1 . The sum of the angles of a triangle is 180°, and the 
sum of the angles of polygon P is 180°. What do you infer 
as to the number of sides of P? The sum of the sides of a 
certain triangle is 180 in., and the sum of the sides of poly¬ 
gon P is 180 in. What do you infer as to the number of 
sides of P? 

State your reasons in both cases, and similarly in Ex. 2. 

2. If three parallels cut off equal segments on one trans¬ 
versal, they cut off equal segments on every other trans¬ 
versal. Three given lines do cut off equal segments on one 
transversal and also cut off equal segments on another 
transversal. What do you infer as to whether or not these 
three lines are parallel ? 

3. The sum of the four sides of any quadrilateral is 
greater than the sum of the two diagonals. 

4. In the fifth century B.C., Hippocrates, a Greek mathe¬ 
matician, proved a theorem which asserts that if three 
semicircles are constructed on the sides 
of a right triangle as diameters, as here 
shown, the crescents L and L' are to¬ 
gether equivalent to the A T. Prove the 
statement. 

This statement is in a somewhat more general form than the one 
given by Hippocrates. 

5. If two altitudes of a triangle are equal, the triangle is 
isosceles; if three altitudes are equal, it is equilateral. 

6. From an external point P a tangent PA is drawn to a 
circle. If the diameter AB and the secant PB, cutting the 
circle at Q, are also drawn, then APAB is similar to A AQP. 

The exercises on pages 272 and 273 have been adapted from various 
examination questions, and represent cases of average difficulty. 




§311 


EXAMINATIONS 


273 


7. Through a point Pinside a circle with center O chords 
whose midpoints are Mi, Mg, Mg, • • • are drawn. Find the 
locus of Ml, Mg, Mg, ‘ 

8. Given a line segment a, construct an equilateral tri¬ 
angle with altitude a. 

9. If the side BA of a AABC is produced through A to 
jD, and if the bisector of ZP meets the bisector of ACAD 
at P, then Z APP = J Z C. 

10. The bisectors of the base A A and P of the equi¬ 
lateral AABC meet in the point P. From P lines are con¬ 
structed II to AC and PC and meeting the base in X and Y 
respectively. Prove that X and Y trisect the base. 

11. Construct a circle which shall have half the area of 
a given circle of radius r. 

12. A circular arch of masonry of radius r feet rests on 

two piers which are d feet apart. Find the height of the 
center of the arch above the level of the top of the piers. 
Discuss the result when r = 25, = 40; when r = 25, cZ = 50. 

13. Without performing the actual construction, show 
how to construct an equilateral triangle equivalent to a 
given square of side s. 

14. A circle of radius 2 in. rolls around the outside of a 
square of side 4 in. Find the length of the path made by 
the center of the circle. 

15. Construct the locus of the center of a circle of radius 
0.5 in. which rolls around an equilateral triangle of altitude 
2 in. Find the length of this locus to the nearest 0.1 in. 

16. While the wind is blowing directly from the north 
at the rate of 10 mi. per hour, a steamer is sailing directly 
east at the same rate. In what direction is a weather vane 
on the ship pointing ? State the reason. 


274 


GENERAL REVIEW 


BOOK V 


Exercises. Optional Trigonometry 

1. Using the right triangle here shown, define, sin A, 
cos A, and tan A in terms of a, b, and c. 

This page is intended only for those who 
have studied trigonometry and are preparing 
for an examination that includes the trigo- 
nometry of the right triangle. The following 
table of natural functions is sufficient for the exercises given below: 


Angle 

sin 

cos 

tan 

Angle 

sin 

cos 

tan 

o 

CO 

0.500 

0.866 

0.577 

60° 

0.866 

0.500 

1.732 

40° 

.643 

.766 

.839 

70° 

.940 

.342 

2.747 

o 

ia 

.766 

.643 

1.192 

00 

.985 

.174 

5.671 



Given the following, find the other parts and the area of 
the right triangle shown above: 

2. A = 30°, a = 20 ft. 5. 5 = 60°, a = 8.2 in. 

3. 40°, 6 = 70 ft. 6. A = 70°, c = 83 yd. 

4. A = 50°, b = 9.5 in. 7. 6 = 20 ft., a = 23.84 ft. 

8. The angle of elevation of a balloon from a point P 
is 60°, and the distance from P to a point directly beneath 
the balloon is 375 yd. Find the height of the balloon. 

9. When a pole 59.6 ft. high casts a horizontal shadow 
50 ft. long, what is the angle of elevation of the sun ? 

10. A fiagpole is broken by the wind, and the upper part 
falls over so as to form a right triangle with the lower 
part and the ground. If the upper part makes an angle of 
70° with the ground and the top of the pole is 15 ft. from 
its foot, find the original height of the pole. 

11. Two sides of a parallelogram are 7 ft. and 9 ft. 6 in. 
respectively, and the included angle is 80°. Find the area 
of the parallelogram. 














SOLID GEOMETRY 


BOOK VI 

LINES AND PLANES IN SPACE 
I. Lines and Planes 

312. Nature of Solid Geometry. In plane geometry we con¬ 
sidered figures lying in a plane, studied their properties 
and relations, and measured their dimensions and areas. 
Such figures are, in general, two-dimensional. 

In solid geometry we shall consider not only figures 
of one dimension and two dimensions, but also three- 
dimensional figures, such as cubes and spheres. 

We shall not need to construct the solid figures by 
means of the straightedge and compasses, and hence we 
shall not discuss any problems of construction. 

313. Plane. A surface such that a straight line joining 
any two of its points lies wholly in the surface is called 
a plane surface, or simply a plane. 



A plane has no thickness, and is understood to be in¬ 
definite in extent. A plane may be conveniently repre¬ 
sented by a thin rectangular solid seen obliquely, as in 
any of the three ways shown above. 

275 






276 


LINES AND PLANES 


BOOK VI 


314. Postulates of Planes. Just as in plane geometry we 
assumed certain postulates upon which to build the proofs 
of the propositions, we now assume certain postulates re¬ 
specting planes. The following are the ones needed in the 
elementary part of solid geometry: 

1. Two intersecting straight lines 
determine a plane. 

Although the plane p may turn about 
one of its lines AB, as shown in the upper 
figure, and occupy any number of positions, 
as p'y p", • • •, it cannot turn if it must also 
pass through an intersecting line CD, as 
shown in the lower figure. In other words, 
the intersecting lines AB and CD deter¬ 
mine the plane p. 

This postulate may be taken to 
include the following statements: 

A straight line and a point not on the line determine a 
plane. 

For example, the line AB and the point C in the second figure are 
sufficient to determine the plane p. 

Three points not in a straight line determine a plane. 

If two of them are connected with the third, we have the case of 
the postulate as first stated. 

Two parallel straight lines determine a plane. 

By definition (§ 51) they must lie in a plane, and one of the par¬ 
allels and any point on the other determine the plane. 

Any one of the four statements given above may be referred to 
as § 314,1. 

2. If two planes have one point in common, they have at 
least one other point in common. 

It is evident that they must then coincide or else that they must 
intersect in a straight line. 









§314 


PLANE DETERMINED 


277 


Exercises. Planes 

1. We commonly say that we live in a space of three 
dimensions, these dimensions being length, width, and 
thickness. We may, then, consider a plane as a space of 
how many and what dimensions? Similarly, a line is a 
space of how many and what dimensions ? 

2. Explain the meaning of the statement that a plane 
passes through a line; that it cuts or intersects the line. 
Draw a figure to illustrate each case. 

3. Two lines in a plane may have no point in common, 
in which case they are parallel; they may have one point 
in common, in which case they intersect; or they may have 
an infinite number of points in common. Write a similar 
statement respecting two planes in three-dimensional space. 

4. Write a statement mentioning three points in the 
room and describing the position of the plane determined 
by them. Illustrate the statement by a drawing. 

5. Write a statement explaining why a three-legged 
stool stands firmly on the fioor while a four-legged chair 
may not do so. 

6. Write a statement describing the position of two 
lines in the room which are so situated that they do not 
determine a plane and do not meet however far produced. 

7. State the geometric reason why a triangle is neces¬ 
sarily a plane figure while a quadrilateral in three-dimen¬ 
sional space need not be. 

8. In three-dimensional space how many different planes, 
are determined by four points ? by five points ? 

9. If n lines, no two of which are parallel, meet a given 
line Z, how many planes are determined, upon what 
postulate does your answer depend ? 


278 


LINES AND PLANES 


BOOK VI 


Proposition 1. Intersection of Planes 

315. Theorem. If two 'planes meet, they intersect in a 
straight line. 



Given two planes p and q which meet. 

Prove that p and q intersect in a st. line. 

Proof. Since p and q meet, they must have at least one 
point, as A, in common. Hence they must have at least one 
other point, as B, in common. § 314, 2 

Draw AB. Post. 1 

Then AB lies in both p and q, § 313 

because otherwise p and q would not he planes. 

Also, no point not on AB can be in both p and q, § 314,1 

because p and q would then coincide instead of meeting. 

Hence the st. line determined by A and B contains all 
points common to p and q. 

. *. AB is the intersection of p and q, 
because this is the meaning of intersection. 

p and q intersect in a st. line. 


Hence 













§§ 315-319 PERPENDICULAR, OBLIQUE, PARALLEL 


279 


316. Perpendicular to a Plane. If a straight line which 
meets a plane is perpendicular to every straight line which 
lies in the plane and passes through the point of meeting, 
the line is said to be perpendicular 
to the plane, and the plane is said 
to be perpendicular to the Him. 

In this figure we may have any num¬ 
ber of planes containing I, and in each 
plane we may have a perpendicular to I 
at O. Hence we may have any number of perpendiculars, as a, b, c, , 
to I at O. 

If, as will be shown (§ 321) to be the case, a,h,c,"- all lie in one 
plane m, then Hs _L to m, and m is ± to 1. 

If we invert the above definition (§ 9), we see that if a 
straight line is perpendicular to a plane, the line is perpen¬ 
dicular to every line in the plane that passes through the 
point of meeting. 

317. Foot of a Perpendicular. The point at which a perpen¬ 
dicular meets a plane is called the foot of the perpendicular. 

318. Oblique to a Plane. If a straight line which meets a 
plane is not perpendicular to the plane, the line is said to 
be oblique to the plane, and the plane is said to be oblique 
to the line. 

Lines which are perpendicular or oblique to a plane are called 
'perpendiculars or obliques respectively. 

When we speak of a perpendicular or an oblique from a point to a 
plane, we mean the line segment from the point to the plane. 

319. Parallel to a Plane. If a straight line cannot meet a 
plane, however far each is produced, the line is said to be 
parallel to the plane, and the plane is said to be parallel to 
the line. Similarly, if one plane cannot meet another plane, 
however far each is produced, the planes are said to be 
parallel. 






280 


LINES AND PLANES 


BOOK VI 


Proposition 2. Perpendicular to a Plane 

320. Theorem. If a line is perpendicular to each of 
two intersecting lines at their point of intersection^ it is 
perpendicular to the plane of the two lines. 



Given AO J- to OP and OR at 0, and m, the plane of OP and OR. 

Prove that ' AO is to m. 

Proof. Through O drav^ any other line OX in the plane 
m, and draw PR, cutting OP, OX, OR in P, Q, and R 


respectively. On AO produced take OA'= OA. 

Join A and A' to P, Q, and R respectively. 

Then AP=A'P and AR=A'R, § 117 

because OP and OR are each ± to AA' at its midpoint. 

.*. A APR is congruent to A A'PR. § 47 

Then ZRPA = ZRPA'. §38 

.*. APQA is congruent to APQA'. § 40 

Then, since AQ = A'Q (§ 38), OQ is _L to AA' at O. § 182 

Hence AO is _L to any line in the plane m through O, 
and thus is J_ to m. § 316 












§§ 320-323 


PERPENDICULARS TO A LINE 


281 


Proposition 3. Perpendiculars to a Line 

321. Theorem. Every line 'perpendicular to a given 
line at a given point lies in a plane perpendicular to 
the given line at the given point 



Given OA, OB, OC, • • • and the plane m, all _L to OY at 0, 

Prove that OA, OB, OC, — ■ lie in m. 

Proof. Suppose that the plane p, determined by OA and 
OY, does not intersect m in OA, but intersects it in OA'. 

Then OA'is_LtoOF. §316 

But OA is J_ to OF. Given 

Hence the supposition is false. 

.p intersects m in OA. Post. 10 

Hence OA lies in m, and similarly for OB, OC,--. § 313 

322. Corollary. Through a given internal point there can 
he one and only one plane perpendicular to a given line. 

323. Corollary. Through a given external point there can 
he one and only one plane perpendicular to a given line. 

In a, the plane of YY' and the given point 
P, let PO be J_ to YY'. In h, any other plane 
containing YY', let OQ be ± to YY'. Then 
m, the plane of OP and OQ, is ± to YY' at O 
(§ 321). Now prove that m is the only X plane 
by using Post. 10. 



















282 


LINES AND PLANES 


BOOK VI 


Proposition 4. Perpendicular through Internal Point 

324. Theorem. Through a given internal 'point there 
can pass one and only one line perpendicular to a plane. 



Given the plane m and the internal point P. 

Prove that through P there can pass one and only one line 
which is _L to m. 

Proof. If a and b are any two lines in the plane m passing 
through P, then through P there is a plane x which is J_ to 


a and a plane y which is _L to 6. § 322 

Since a and h are not identical but meet at P, x and y are 
not identical and must intersect in a st. line BQ. § 315 
Since a is _L to x^ it is _L to PQ. § 316 

Similarly, 6 is _L to PQ, 

and hence PQ is _L to m. § 320 

Now if another linePQ' could pass through Pand be J_ 
to m, it would be ± to a and to 6. § 316 

Hence PQ' would lie in both x and y, § 316 

and PQ' would coincide with PQ. 

. *. PQ is the one and only _L to m through P. 











§§ 324, 325 


PERPENDICULAR TO A PLANE 


283 


Proposition 5. Perpendicular through External Point 

325. Theorem. Through a given external point there can 
pass one and only one line perpendicular to a plane. 



Given the plane m and the external point P. 

Prove that through P there can pass one and only one line 
which is _L to m. 

Proof. Let I be any line in m, and let PA be J_ to 1. 

In m construct AR _L to I, and in the plane of AR and P 
construct PO-L to Ai?. §§104,105 

Produce PO to P', making OP' = OP. Post. 2 

Let OB be any other line from O to I, and draw PP, 
P'A, and P'P. 

Then Z is J- to plane AP'P. § 316 

Now prove that rt. A APB is congruent to rt. AAP'B, hence that 
A OPB is congruent to A OP'B, and hence that PO is _L to OB. 

Then PO is J_ to m. § 316 

Further, if PQ is any other line from P to m and QO is 
drawn, then Z.QOP is a rt. Z. § 316 

Hence PQ is not ± to m, and PO is the only ±. 











284 


LINES AND PLANES 


BOOK VI 


Proposition 6. A Perpendicular and Obliques 

326. Theorem. If from an external 'point a perpen¬ 
dicular and obliques are dra'wn to a plane^ 

1. The perpendicular is shorter than an'y oblique; 

2. Obliques meeting the plane at equal distances from 
the foot of the perpendicular are equal; 

3. Of two obliques meeting the plane at unequal dis¬ 
tances from the foot of the perpendicular, the more 
remote is the longer. 



Given the plane m, the external point P, PO X to and the 
obliques PA, PB, PC drawn to m so that OA > OB = OC. 

Prove that PO < PC, PB = PC, and PA > PC. 

Proof. Produce PO to P', making OP' = OP. Post. 2 

Draw P'C. Post. 1 

Now prove that PP'<PC+CP', and hence that PO<PC. 

Then prove that A OBP is congruent to A OOP, 
and hence that PB = PC^ 

Finally, prove that PA>PB, 

and hence that PA>PC. 













§§326-329 PERPENDICULARS AND OBLIQUES 


285 


327. Distance. The length of the perpendicular from a 
point to a plane is called the distance from the point to 
the plane. 

The following corollaries (§§ 328, 329) extend the idea of a locus with 
which the student is familiar from plane geometry. 

328. Corollary. The locus of 'points equidistant from the 
vertices of a triangle is a line through the center of the cir¬ 
cumscribed circle, and perpen¬ 
dicular to the plane of the 
triangle. 

We have first to prove (§ 181) 
that any point P on the A.OY satis¬ 
fies the conditions; that is, that 
AP — BP = CP. But this follows 
from §§ 134,1 and 326, 2. 

We have then to prove that any 
point P which satisfies the conditions 
that AP = BP = CP is on the line OY. Now AO = BO = CO (§ 134,1). 
Then since OP is a common side, A A OP, POP, and OOP are congruent 
(§ 47). Then the A made by OP with any lines in m are equal, and hence 
they are rt. A. Hence OP is JL to m at O (§ 316), and since OP is part 
of OP(§ 324), Pis on OY. 

329. Corollary. The locus of points equidistant from two 
given points is the plane perpendicular at the midpoint to 
the line segment joining the points. 

We have first to prove (§ 181) that any 
point P in m satisfies the condition that 
PA = PB. We have then to prove that any 
point P' such that P'A = P'B lies in m, 
which is best done by an indirect proof. 

The proof of each of these steps is left 
for the student. 

We here meet a case in which the locus 
is a plane instead of a line. In plane geometry, as the student has 
already found, a locus is usually a line; in solid geometry a locus may 
be a line or it may be a surface. 



Y 











286 


LINES AND PLANES 


BOOK VI 


Exercises. Lines and Planes 

1. Equal oblique lines drawn from a point to a plane 
meet the plane at equal distances from the foot of the per¬ 
pendicular from the point to the plane; and of two unequal 
oblique lines the greater meets the plane at the greater 
distance from the foot of the perpendicular. 

2. The locus of points equidistant from all points on a 
circle is a line through the center, perpendicular to the 
plane of the circle. 

3. Find the locus of points at a given distance from each 
of two given points. 

4. Explain how a carpenter might proceed to set a joist 
so that it shall be perpendicular to a horizontal floor. Draw 
a flgure to illustrate any method which seems practical to 
you for the carpenter to use. 

5. What geometric principle is involved in the statement 
that if the spoke of a wheel is perpendicular to the axle, 
the spoke determines a plane as the wheel revolves ? 

6. A steel smokestack 80 ft. high is braced by four steel 
wires each 100 ft. long and reaching from the top of the 
stack to the ground. If the wires are straight, at what dis¬ 
tance from the foot of the stack does each reach the ground? 
Would three wires serve as well ? Would two serve as well ? 
Would one serve as well? State the geometric principle 
involved in each case. 

7. Through a point P there pass four lines such that no 
three are in the same plane. Find the number of planes 
determined by the four lines. 

8. In a plane P there lie four lines such that no three 
pass through the same point. Find the greatest number 
of points that can be determined by the four lines. 


§§ 330-332 


PARALLEL LINES 


287 


Proposition 7. Perpendiculars to a Plane 

330. Theorem. Two lines 'perpendicular to the same 
plane are parallel. 



Given AB and CZ>, each JL to the plane m. 

Prove that AB is W to CD. 


Proof. Draw AD and BD, and in m draw PQ _L to BD at 
Z), making DP=DQ. Draw AP, AQ, BP, BQ. 


By congruent A (§ 47) prove that A ADP = ZADQ = 90°, and then 
that BD, CD, and AD lie in the same plane (§ 321). Then prove that 
AB also lies in this plane (§ 313), and then that AB and CD are each ± 
to BD{% 316). 


, AB is II to CD. 


§57 


331. Corollary. If one of two parallel lines is perpendicular 
to a plane, the other is also perpendicular 
to the plane. 

For if through any point O of 6 a line is drawn X 
to m, how is it related to a (§ 330)? Now apply § 52. 

332. Corollary. If two lines are parallel 
to a third line, they are parallel to each 
other. 

For if 6 is X to m, so are a and c (§ 331).. 






















288 


LINES AND PLANES 


BOOK VI 


Proposition 8. Parallel Lines 

333. Theorem. If two lines are parallel^ every plane 
containing one and only one of the . lines is parallel to 
the other. 



Given the II lines AB and CD, and the plane m containing 
CD but not AB. 


Prove that m is II to AB. 

Proof. AB and CD determine a plane n, § 314, 1 
and AB lies in n, however far each is produced. § 313 

Hence, if AB meets m, it meets CD. § 313 

Since AB cannot meet CD, § 51 

AB cannot meet m; 

that is, m is II to AB. § 319 


334. Corollary. Through either of two lines not in the 
same plane one and only one plane can 
pass parallel to the other line. 

For if AB and CD are the given lines, and 
if CX is II to AB, what can be said of the 
plane m, which is determined by CD and CX, 
with respect to the line AB ? Why can there be only one such plane ? 

Lines placed like AB and CD in this figure are called skew lines. 

















§§ 333-335 


PARALLEL LINES 


289 


335. Corollary. Through a given point one and only one 
plane can pass parallel to each of two given lines not in the 
same plane. 

Let P be the given point and AB and CD the given lines. If, now, 
we construct through P the line A'B' II to AB, and the line C'D' II to 
CD, these lines determine the plane m. 



Then prove that m is II to AB and CD, 
and that no other such plane is possible 
through P. 


In the above figure, the lines AB and 


CD are said to form an angle, although X ^ / 

they do not meet. This angle is defined 

as the Z. C'PB', but the concept is rarely used in elementary geometry. 


Exercises. Lines and Planes 


1. State the geometric principle by which we know that 
a straight edge results from folding a piece of paper. 

2. In a given plane what is the locus of points equi¬ 
distant from two parallel lines in the plane? Given two 
parallel planes instead of two parallel lines, what is the cor¬ 
responding locus in a space of three dimensions? Draw 
the figures but give no proofs. 

3. If a given line is parallel to a given plane, the inter¬ 
section of the plane with any plane passed through the 
given line is parallel to that line. 

4. If a given line is parallel to a given plane, a line parallel 
to the given line drawn through any point of the plane lies 
in the plane. 

5. If equal oblique lines are drawn from a given external 
point to a plane, they make equal angles with lines drawn 
from the points where the oblique lines meet the plane to 
the foot of the perpendicular drawn from the given point 
to the plane. 










290 


LINES AND PLANES 


BOOK VI 


Proposition 9. Parallel Planes 


336. Theorem. Two planes perpendicular to the same 
line are parallel. 



Given the planes m and n, each -L to the line /. 

Prove that m is II to n. 

Proof. If m is not II to n, it must meet n, in which case 
we should have two planes through a point in their inter¬ 
section both J_ to 1. 

Since this is impossible (§ 322), m is II to n. 

The following corollary, which is analogous to the Postulate of 
Parallels (§ 52), may be assumed, if desired, without proof. 


337. Corollary. Through a given external point one and 
only one plane can pass parallel to a given plane. 

If P is the point and m is the plane, as shown in the left-hand 
figure, there is only one line PQ that is ± to m (§ 325). Through P 
there is one and only one plane 
n that is _L to PQ (§ 322), and 
this is II to w (§ 336). 

If through P there were two 
planes II to m, one would be 
oblique to PQ, as shown in the 
right-hand figure, and would con¬ 
tain some line PP' that would meet its projection QQ' in m. Then n 
would not be II to m (§ 319). Hence only one plane through P is II to m. 


























§§ 336, 337 


REVIEW EXERCISES 


291 


Exercises. Review 

1. If from the foot of a perpendicular to a plane a line 
is constructed at right angles to any line in the plane, the 
line drawn from its intersection with the line in the plane 
to any point on the perpendicular is perpendicular to the 
line in the plane. 

2. If two perpendiculars extend from a given external 
point to a plane and to a line in that plane respectively, the 
line joining the feet of the two perpendiculars is perpen¬ 
dicular to the given line. 

3. From two vertices of a triangle perpendiculars are 
constructed upon the opposite sides. From the intersection 
of these perpendiculars there is a perpendicular to the plane 
of the triangle. Prove that a line drawn to any vertex of 
the triangle from any point on this perpendicular is perpen¬ 
dicular to the line drawn through that vertex parallel to 
the opposite side. 

4. Find the point in a plane to which lines may be drawn 
from two given external points on the same side of the 
plane so that their sum shall be the least possible. 

From one point A suppose that a line O is ± to the plane and that 
it is produced to A', making OA'= OA. Connect A' and the other point B 
by a line cutting the plane at P. Then AP APB is the least sum. 

5. If three equal oblique lines are drawn from an exter¬ 
nal point to a plane, the perpendicular from the point to 
the plane meets the plane at the center of the circle circum¬ 
scribed about the triangle which has for its vertices the 
points where the oblique lines meet the plane. 

6. State and prove the propositions of plane geometry 
corresponding to §§330, 331, and 332. Why do not the 
proofs of those propositions apply to the corresponding 
propositions of solid geometry? 


292 


LINES AND PLANES 


BOOK VI 


Proposition 10. Parallel Planes Intersected 

338. Theorem. If two parallel planes are cut by a third 
plane^ the lines of intersection are parallel. 



Given the II planes m and n, intersected by a third plane p 


in AB and CD respectively. 

Prove that AB is II to CD. 

Proof. AB and CD are in the same plane p. Given 
Since AB is always in m and CD is always in n, § 313 

m and n must meet if AB and CD meet. 

But m is II to n, Given 

and hence m and n cannot meet. § 319 

.*. A5is II toCD. §51 


339. Corollary. A line perpendicular to one of two parallel 
planes is perpendicular to the other 
also. 

Let PQ be ± to m, and let p and q be two 
planes containing PQ. Now prove (§ 338) that 
QB is II to PA and that QD is II to PC. Then 
prove that PQ is ± to QB and QD, and hence 
that PQ is ± to n. 





















§§ 338-342 


PARALLEL PLANES 


293 


340. Distance between Parallel Planes. The length of a 
perpendicular line segment between two parallel planes is 
called the distance between the planes. 

It has been shown (§ 339) that if this segment is perpendicular to 
one of two parallel planes it is perpendicular to the other. It will now 
be shown (§ 341) that the length is the same whatever perpendicular 
between the planes is taken. 

341. Corollary. Two parallel planes are everywhere equi¬ 
distant from each other. 

In the figure of §339, if AB is constructed II to PQ, then ABQPis 
a O (§ 72). Since it is given that PQ is J. to m, then AB is also ± to m 
(§ 331). Both AB and PQ are then ± to n (§ 339) and represent distances 
measured on any two _h. But AR = PQ(§ 76), and hence m and n are 
everywhere equidistant from each other. 

342. Parallel in the Same Sense. If two parallel rays lie 
on the same side of the line segment which joins their 
end points, they are said to be parallel in the same seme. 

Exercises. Parallel Planes 

1. Parallel lines included between parallel planes are 
equal. 

2. The locus of points equidistant from two parallel 
planes is a plane which is perpendicular to a line perpen¬ 
dicular to the planes and which bisects the segment cut 
off by them. 

3. The locus of points equidistant from two parallel lines 
is a plane which is perpendicular to a line perpendicular 
to the given lines and which bisects the segment cut off 
by them. 

4. The locus of points at a given distance from a plane 
is a pair of parallel planes, each at the given distance from 
the given plane. 


294 


LINES AND PLANES 


BOOK VI 


Proposition 11. Arms of Angles Parallel 

343. Theorem. If two angles not in the same plane 
have their arms respectively parallel in the same sense, 
the angles are equal and their planes are parallel. 



Given the A A and A' in the planes m and n respectively, 
with their arms respectively II in the same sense. 

Prove that AA = AA' and that m is W to n. 

Proof. Take AB=A'B\ and AC—A'C\ and draw BC, 
B'C\ AA', BB\ CC\ 

Then A A' is equal and II to BB' and to CC', § 81 

Hence BB'=CC'{Kx. 5), and BB' is II to CC'(§ 332). 

Since5C=5'C'(§76), AA^Ciscongruent toAA'^'C'. § 47 
.•.ZA = ZA'. §38 

Now if m is not II to n, they will meet in a line 1. § 315 

Since AB and AC are II to ?^ (§ 333), neither can meet 1. 

But since both AB and AC cannot be II to Z (§ 332), m can¬ 
not meet n, and hence m is II to n, 

344. Corollary. If two intersecting lines are each parallel 
to a plane, the plane of these lines is parallel to that plane. 

In the figure of § 343, if AB and AC are both II to n, show that AB 
cannot meet a line A'B', which is the intersection of n and the plane 
of AA' and AB. Similarly, AC is II to A'C'. Hence m is II to n (§ 343). 















§§ 34Sf-345 


TRANSVERSALS IN SPACE 


295 


Proposition 12. Transversals in Space 


345. Theorem. If two lines are cut hy three parallel 
planeSy their corresponding segments are proportional. 



Af Ej By and C, F, D respectively. 


Prove that 


AE_CF 
EB Fd' 


Proof. Let q intersect the plane of A, B, D in EG and 
the plane of A, D, C in GF. 


Then 

EG is II to BD and GF is II to AC. 

§338 

Hence 

AE 

AG 

gd' 


EB 


and 

CF_ 

AG 

§201 

FD 

gd' 


Hence 

AE_ 

CF 

Ax. 5 

EB 

fd' 



It should be observed that this proposition is a generalization of § 203, 
which applies only to a figure lying in one plane. It may be stated 
still more generally as follows: If two lines are cut hy any number of 
parallel planes^ their corresponding segments are proportional. 

Consider also the case of AB intersecting CD between the planes. 
















296 


LINES AND PLANES 


BOOK VI 


Exercises. Review 

1. In a given plane find the locus of points equidistant 
from two given points not in the plane. 

2. Find the locus of points equidistant from three given 
points not in a straight line. 

3. Find the locus of points equidistant from two given 
parallel planes and also equidistant from two given points. 

4. What is the locus of points at a given distance from 
each of two planes ? 

5. The line AB cuts three parallel planes in the points A, 
E, B ; and the line CD cuts these planes in the points C, F, D. 
If AE=Z in., EB — A in., and CD = 6 in., what are the lengths 
of CFandFD? 

6. In Ex. 5, if AB = 1^ in., 0^ = 10 in., and CD = 18 in., 
what are the lengths of AE and EB^. 

7. It is proved in plane geometry that if three or more 
parallels intercept equal segments on one transversal, they 
intercept equal segments on every transversal. State and 
prove a corresponding proposition in solid geometry. 

8. It is proved in plane geometry that the line which 
joins the midpoints of two sides of a triangle is parallel 
to the third side. State and prove a proposition in solid 
geometry which shall refer to a plane passing through the 
midpoints of two sides of a triangle. 

9. A cylindric water tank which is 16 ft. deep and 12 ft. 
in diameter is filled with water to a depth of 9 ft. A pole 
standing obliquely in the tank just reaches from a point 
on the circumference of the base to a point exactly oppo¬ 
site on the upper rim. Find the length of that part of the 
pole which is under water. 

10. Consider Ex. 9 when the water level rises 3 ft. 


§§ 346-350 


DIHEDRAL ANGLES 


297 


II. Dihedral Angles 


346. Half-Planes. Any straight line in a plane is said to 
divide the plane into two half-planes. 

This term corresponds to the term rays in plane geometry. 

347. Dihedral Angle. If two half-planes proceed from the 
same line, they form a dihedral angle. 

In this figure the half-planes p and q 
are the faces of the dihedral angle, and AB 
is the edge. 

This dihedral angle may be designated 
by pq, d, p-AB-q, or AB, of which the first 
two forms are the most convenient. 



348. Plane Angle of a Dihedral Angle. A plane angle whose 
arms are perpendicular to the edge of a 
dihedral angle and lie respectively in the 
faces is called the plane angle of the 
dihedral angle. 

For example, if OA, OB] O'A', O'B'; 0"A", 

0"B" are each perpendicular to 00' in this figure, 
each of the AO, O', O" may be taken as the plane 
angle of the dihedral angle. 

349. Corollary. Two dihedral angles have the same ratio 
as their plane angles. 

For it is evident that the amount of turning necessary to generate 
a dihedral angle is the same as that which is necessary to generate its 
plane angle. Hence their numerical measures are always identical. 

If a proof were necessary, it would be substantially the same as 
the one in § 136. 

350. Kinds of Dihedral Angles. A dihedral angle is right, 
acute, or obtuse according as its plane angle is right, acute, 
or obtuse. 

Similarly we may use the terms straight, vertical, adjacent, comple¬ 
mentary, supplementary, and oblique in connection with dihedral angles. 

PS 









298 


DIHEDRAL ANGLES 


BOOK VI 


351. Relation of Dihedral Angles to Plane Angles. It is 
evident that the proofs of many properties of dihedral 
angles are identical with those of analogous properties of 
plane angles. A few of the more important propositions 
will be proved, but the following may be assumed without 
proof or may be taken as exercises: 

1. If a plane meets another plane, it forms with it two 
adjacent dihedral angles whose sum is equal to two right 
dihedral angles. 

2. If the sum of two adjacent dihedral angles is equal to 
two right dihedral angles, their exterior faces are in the 
same plane. 

3. If two planes intersect each other, the vertical dihedral 
angles are equal. 

A. If a plane intersects two parallel planes, the alter¬ 
nate dihedral angles are equal; the corresponding dihedral 
angles are equal; and the two interior dihedral angles on 
the same side of the transverse plane are supplementary. 

5. When two planes are cut by a third plane, if the 
alternate dihedral angles are equal, or the corresponding 
dihedral angles are equal, and the edges of the dihedral 
angles thus formed are parallel, the two planes are parallel. 

6. Two dihedral angles whose faces are parallel each to 
each are either equal or supplementary. 

7. Two planes parallel to the same plane are parallel to 
each other. 

While the proofs of the above propositions are valuable as exercises 
in logic, they are not essential. 

352. Perpendicular Planes. If two intersecting planes 
form a right dihedral angle, the planes are said to be 
perpendicular to each other. 

As in all definitions we may invert the statement and say that, if the 
planes are perpendicular, they form a right dihedral angle. 


§§ 351-355 


PERPENDICULAR PLANES 


299 


Proposition 13. Perpendicular Planes 

353. Theorem. If two planes are perpendicular to each 
othery a line drawn in one of them perpendicular to their 
intersection is perpendicular to the other. 



Given the planes m and n -L to each other, and the line CD 
in m _L to ABy the intersection of m and n. 


Prove that 

CD is ± to n. 


Proof. In n construct DE J_ to AB at D. 

§104 

Then 

Z.EDC is a rt. Z. 

§352 

Also, 

Z.CDA is a rt. Z. 

Given 


. *. CD is Z to n. 

§320 


354. Corollary. If two planes are perpendicular to each 
othery a line perpendicular to one of them at any point of 
their intersection lies in the other. 

In the figure of § 353, let DC be constructed in the plane m ± to AB 
at any point D. Then DC is _L to n (§ 353). Now through D there can 
be only one _L to (§ 324). Hence a ± to n through D must coincide 
with DC and lie in the plane m. 

355. Corollary. If two planes are perpendicular to each 
other, a line perpendicular to one plane through any point 
in the other lies in the second plane. 

In the figure of § 353, if CD is constructed in the plane m J. to AB 
through C, it is J. to n (§ 353). But only one such ± is possible (§ 325). 















300 


DIHEDRAL ANGLES 


BOOK VI 


Proposition 14. Plane through a Perpendicular 

356. Theorem. If a line is 'perpendicular to a planey 
every plane passed through this line is perpendicular 
to the plane. 



Given the line CD _L to the plane n at the point D, and any 
plane m through CD. 

Prove that mis 1. to n. 

Proof. Let AB be the intersection of the planes m and n. 


In n construct DE ± to AB at D. § 104 

Since CD is J_ to n, Given 

then CD is ± to AB. § 316 

If a st. line is A. to a plane, the line is J_ to every line in the 
plane that passes through the point of meeting. 

Now Z.EDC measures the dihedral Znm, § 349 

because the plane Z. of a dihedral Z may he taken as the 
measure of the dihedral Z. 

But ZlEDC is a rt. Z, § 316 

because CD is given Zto n. 


. *. m is X to 


§352 














§§ 356,357 PLANE THROUGH A PERPENDICULAR 


301 


357. Corollary. Through a line not perpendicular to a 
given plane, one and only one plane can pass perpendicular 
to the given plane. 

Let AB be any line oblique to plane m, 
and from any point P on AB let a ± PQ be 
drawn to m. Then the plane n, determined 
by AB and PQ, is _L to m (§ 356), so that one 
such plane is possible. 

If another such plane, say x, were possi¬ 
ble, PQ would lie in x (§ 355), so that x would be determined by AB and 
PQ, and hence would coincide with n. 



Exercises. Planes and Perpendiculars 

1. A plane perpendicular to the edge of a dihedral angle 
is perpendicular to each of its faces. 

2. If two intersecting planes are respectively perpen¬ 
dicular to two intersecting lines, the line which is deter¬ 
mined by the planes is perpendicular to the plane which is 
determined by the lines. 

3. .Consider Ex. 2 after interchanging the words'' planes ^ ^ 
and 'Hines'’ in every instance. 

4. The plane passing through a given point P and per¬ 
pendicular to the edge of a given dihedral angle contains 
the perpendiculars from P to the faces of the angle. 

5. A workman with a 12-foot pole, a piece of chalk, and 
a string stands in a room that is 10 ft. high. By the use 
of these instruments how can he find a point on the floor 
that is directly below a given point on the ceiling ? 

6. If two planes are perpendicular to each other, a line 
perpendicular to either plane through an internal point not 
on their intersection is parallel to the other plane. 

7. Consider Ex. 6 after substituting the word "plane" 
for "line" and "line" for "plane." 









302 


DIHEDRAL ANGLES 


BOOK VI 


Proposition 15. Two Perpendicular Planes 

358. Theorem. If two intersecting planes are each per¬ 
pendicular to a third plane, their intersection is also 
perpendicular to that plane. 



Given the planes m and n intersecting in AB and each J_ to 
the plane q. 

Prove that AB is _L to q. 

Proof. Through P, any point on AB, there can pass but 
one line, as PX, which is _L to g. § 325 

Now PX must lie in both m and n, § 355 

Hence PX coincides with AP (§ 315); that is, AB is J_ to q. 

359. Corollary. The locus of points equidistant from two 
intersecting planes is the pair of planes which bisect the 
dihedral angles formed hy the given 
planes. 

Let plane h bisect the dihedral Anm. 

We must first show that, if the JsPQ, PR 
from any point P to m and n are equal, P lies 
in 6. Now g, the plane of PQ and PR, is ± to 
m and n (§ 356). Then OA is ± to g (§ 358). 

Since A OPQ is congruent to A OPR (§ 71), 

ZPOQ = ZPOR (§ 38). Then P lies in 6. 

Now prove by congruent A (§ 68) that for any point P lying in 6, PQ = PR. 

A similar proof evidently holds for the plane a. 




















§§ 358,359 


REVIEW EXERCISES 


303 


Exercises. Review 

1. If three equal lines are drawn to a plane from an ex¬ 
ternal point, the perpendicular from the point to the plane 
determines the center of the circle circumscribed about the 
triangle determined by the intersections of the planes of 
the three lines with the given plane. 

2. If three lines not in the same plane meet in a point, 
how shall a line be drawn so as to make equal angles with 
all three of these lines ? 

After proving Exs. 3-5, interchange the words ^Hine” and 
'' plane '' and prove the resulting statements: 

3. If a line is parallel to one of two parallel planes, it is 
also parallel to the other plane. 

Exs. 3-5 illustrate what are known as dual propositions, which are 
propositions that are also true if we interchange such terms as 'Mine” 
and "plane.” Another kind of dual propositions is seen in Exs. 6 and 7. 

4. If a plane and a line not in the plane are perpen¬ 
dicular to the same line, they are parallel. 

'' In the plane ’ ’ will become '' through the line' ’ in the dual. 

5. Parallel planes make equal angles with a given line. 

If P is an external point and PA is a segment from P to a plane, let 

PQ be perpendicular to the plane. Then the inclination of PA to the 
plane, or the angle which it makes with the plane, is the Z QAP. 

In Exs. 6 and 7 interchange ^'poinV^ and "^plane^^ and 
draw figures to illustrate each statement and its dual: 

6. Three points, in general, determine a plane. 

7. A point and a line determine a plane. 

8. If a plane is perpendicular to one of two parallel lines, 
it is perpendicular to the other line also. Write the dual, 
as in Exs. 3-5, and prove it. 


304 


DIHEDRAL ANGLES 


BOOK VI 


Proposition 16. Common Perpendicular 

360. Theorem. Between two lines not in the same plane 
there is one and only one common perpendicular. 



Given AB and CD, two lines not in the same plane. 

Prove that there is one and only one common _L between 
AB and CD, 


Proof. 

In the plane of A and CD let AX be II to CD, § 107 

Then 

m, the plane of AX and AB, is II to CD, 

§333 

Let 

DD' be the ± from D to m. 

§325 

Then 

n, the plane of DD' and CD, is _L to m. 

§356 

Let 

n intersect m in C'D', 


Then 

CD cannot meet C'D' without meeting m. 

§313 

But 

m is II to CD, 

Proved 


.*. CD is II to C'D', 

§51 

If 

AB is II to C'D', 



AB is also II to CD, 

§332 

Since 

AB and CD are not in the same plane, 

Given 


AB is not II to CD, 

§51 


.*. AB must intersect C'D', 



Designate the point of intersection as P. 














§360 

COMMON PERPENDICULAR 

305 

Let 

PQ be -L to m. 

§324 

Then 

PQ is X to AB and to C'D', 

§316 

and hence 

PQ is X to CD. 

§63 

Hence 

If there 

there is one common J_. 
is another common _L, suppose it to be EA. 

Then 

EA is _L to AX, 

§63 

and hence 

EA is J_ to m. 

§320 

Let 

EE'he _L to C'D'. 

§ 105 

Then 

EE' is _L to m. 

§353 

But if 

EA is ± to m, 

EE' cannot be J_ to m. 

§325 


That is, the supposition that there is a second common 
-L, as EA, leads to an impossible result. 

Hence there is one and only one common J_ between the 
lines AB and CD. 


Exercises. Review 

1. The common perpendicular between two lines not in 
the same plane is the shortest line joining them. 

2. If three lines passing through a given point P are cut 
by a fourth line that does not pass through P, the four lines 
all lie in the same plane. 

3. If seven lines, no three of which lie in the same plane, 
pass through the same point, how many planes are deter¬ 
mined by these lines ? 

4. A cubic tank 10 in. deep is filled with water to a depth 
of 7 in. A foot rule resting on and oblique to the bottom just 
reaches the top edge of the tank. Make a sketch of the 
tank, and compute the length of the rule covered by water. 

5. A plane perpendicular to one of two parallel planes 
is also perpendicular to the other plane. 


306 


DIHEDRAL ANGLES 


BOOK VI 


6. If the walls of a room are perpendicular to both the 
ceiling and the floor, what is the geometric reason for 
asserting that the ceiling is parallel to the floor ? 

7. If four points lie in a straight line, can their pro¬ 
jections on a plane lie in a straight line? Must they lie 
in a straight line? Draw the flgures to illustrate your 
answers and state the geometric principles involved. 

8. In plane geometry we find that three lines, in gen¬ 
eral, determine three points; namely, the vertices of the 
triangle formed by the lines. Draw a figure illustrating 
the corresponding case for planes in solid geometry and 
state what is determined. 

9. The equal sides of an isosceles triangle make equal 
angles with any plane that contains the base. 

10. The base BC of the isosceles A ARC in the plane m 
is 3 in., and the perimeter of the triangle is 10 in. If the 
triangle revolves about its base as an axis, what is the 
greatest distance, to the nearest 0.001 in., from the plane 
that is reached by A? 

In finding roots or powers, the student should make use of the table 
given on page 462. 

11. A point P moves so as to be constantly 10 in. from 
each of the points A and B, which are 8 in. apart. Find to 
the nearest 0.001 in. the length of the locus of P. 

12. Two parallel planes m and n are cut by a third plane P 
so that one of the dihedral angles contains 32° 45'. Find 
the sizes of the other dihedral angles. 

13. From a point P which is 12 in. above a plane m, PO is 
drawn perpendicular to m. With 0 as center and a radius 
of 9 in. a circle is drawn in m. At Q, any point on this 
circle, a tangent QR, 20 in. in length, is drawn in m. Find 
the length of PR to the nearest 0.01 in. 


§§ 361,362 


POLYHEDRAL ANGLES 


307 


III. Polyhedral Angles 

361. Polyhedral Angle. If three or more planes meet in a 
point, they form a polyhedral angle. 

If we consider angles greater than 360°, two lines which meet form 
an infinite number of plane angles; but this 
fact never confuses us in plane geometry. 

Similarly, in solid geometry, if three planes 
meet in a point, they form an infinite num¬ 
ber of polyhedral angles; but since we shall 
always make clear the angle to be considered, 
this fact will cause no difficulty. 

The three planes AVB, BVC, and CVA, in the above figure, form a 
polyhedral angle which we designate by V, or by V-ABC. The letters 
are given in the order in which they occur 
around the figure. 

If the figure formed on a plane which cuts 
all the planes forming a polyhedral angle is 
convex, the angle is said to be convex ; if the 
figure is concave, as here shown, the angle is v 
said to be concave. Since we shall consider 
only convex polyhedral angles, this distinction need not be memorized. 

362. Parts of a Polyhedral Angle. The common point at 
which the planes meet to form a polyhedral angle is called 
the vertex of the angle. The intersections of the planes are 
called the edges of the angle. The portions of the planes 
lying between the edges are called the faces of the angle. 
The angles formed by adjacent edges are called the face 
angles of the polyhedral angle. The vertex, edges, faces, 
face angles, and the dihedral angles formed by the faces 
are the parts of a polyhedral angle. 

In a polyhedral angle of n faces there are n edges, 
n face angles, and n dihedral angles. 

In the first of the above figures, the vertex is V ; the edges are VA, 
VB,VC; the faces are AVB, BVC, CVA; the face angles are ZAVB, 
ZBVC, ZAVC; and the dihedral angles are VA, VB, VC. 





308 


POLYHEDRAL ANGLES 


BOOK VI 


363. Classes of Polyhedral Angles. A polyhedral angle of 
three faces is called a trihedral angle. 

Polyhedral angles of four, five, six, seven, • • • faces take the prefixes 
tetra-, penta-, hexa-, hepta-, • • •. Since, however, we rarely refer to 
polyhedral angles of more than three faces, these names need not be 
memorized. 

364. Equal Polyhedral Angles. If two polyhedral angles . 
can be placed so that their ver¬ 
tices and edges coincide, the 
angles are said to be equal. 

Conversely, if polyhedral angles are 
equal, they can be made to coincide by 
superposition, because, if the vertices ^ 
and edges coincide, all the correspond¬ 
ing parts coincide. In this figure the trihedral A V and V' are equal. 



Exercises. Review 


1. A reading lamp is attached to an upright rod which 
is fastened to two iron pieces, or feet, resting on the floor 
as here shown. If the rod is perpendicular to 
the two pieces, is it perpendicular to the floor ? 

Would three pieces be better ? Would four be 
better? Would five be still better? State the 
geometric principles involved in your answers. 




2. Two adjacent walls and the ceiling of a _ JL _ 
rectangular room form a trihedral angle. Write 
a statement of the relations of the parts of the angle; for 
example, that each dihedral angle has a certain size. 


3. It is known that the projections of four points upon 
a plane (that is, the feet of the perpendiculars from the 
points to the plane) lie in a straight line. Write your con¬ 
clusion as to Vhether or not these points lie in a straight 
line; lie in a plane; are scattered at random in space. 






§§ 363-366 


SYMMETRIC ANGLES 


309 


365. Symmetric Polyhedral Angles. If the faces of the 
polyhedral ^V-ABCD are produced through the vertex 
y, another polyhedral angle, the ZV-A'B'C'D', is formed. 
The ^V-A'B'C'D' is said to be 
symmetric with respect to 
ZV-ABCD, 

The face AAVB, BVC, - •, in the 
figure at the left, are’ equal respec¬ 
tively to the face AA'VB', B'VC, • • • 
of the polyhedral ZV-A'B'C'D' 

Also, the dihedral AVA, VB, • • • are 
equal respectively to the dihedral 
A VA', VB', • • • (§ 351, 3). The figure at the right shows a pair of these 
vertical dihedral angles. 


B' B' 



Looked at from the point V, the edges of Z.V-ABCD are 
arranged counterclockwise (from left to right) in the order 
VA, VB, VC, VD, but the edges of ZV-A'B'C'D' are arranged 
clockwise (from right to left) in the order VA', VB', VC', VD'; 
that is, in an order which is the reverse of the order of the 
edges of /LV-ABCD, Therefore, 

Two symmetric polyhedral angles have all their parts equal, 
each to each, hut arranged in reverse order. 


366. Symmetric Polyhedral Angles not Superposable. In gen¬ 
eral, two symmetric polyhedral angles cannot be superposed. 

If the trihedral Z.V-A'B'C' here shown is . 
made to turn 180° about XY, the bisector of 
Z.CVA', then VA' coincides with VC, yc'with 
yA, and the face A'VC' with the faceAyC. 

But since the dihedral ZyA, and hence the 
dihedral ZyA', is not equal to the dihedral 
ZyC, the face A'VB' does not coincide with 
the face BVC, nor does C'ys' with A VB. Hence 
VB' takes some position as VB" ; that is, symmetric trihedral 
angles cannot, in general, be superposed. 




310 


POLYHEDRAL ANGLES 


BOOK VI 


Proposition 17. Trihedral Angles 

367. Theorem. If the three face angles of one trihe¬ 
dral angle are equal respectively to the three face angles 
of another^ the trihedral angles are either equal or 
symmetric. 



Given the trihedral and V^-X'Y'Z' with face 

AYVX, ZVXy ZVY equal respectively to face AY'VX'y 
Z'VX'y Z^V’Y'. 

Prove that the trihedral A V-XYZand V'-X'Y'Z' are either 
equal or symmetric. 

Proof. On the edges of the trihedral A take the six equal 
segments VA, VB, VC, V'A', V'B', V'C'. 


Draw 

AB, BC, CA, A'B\ B'C', C'A\ 


Since 

the face A are respectively equal, 

Given 

then 

/\BAV is congruent to /\B'A'V\ 

A CAP is congruent to AC'A'y', 


and 

ACRF is congruent to /\C'B'V\ 

§40 

Hence 

AB=A'B\ BC=B'C\ CA=C'A\ 

§38 


ABAC is congruent to AB'A'C'. 

§47 











§367 


TRIHEDRAL ANGLES 


311 


From any point P on VA construct PQ in the face XVY 
and PR in the face XVZ, each _L to VA. § 104 

Since zi VABand VACare equal A of isosceles A, 

A VAB and VAC are acute. § 65 

Hence the _k PQ and PR meet AB and AC respectively. 
Draw QR. 

OnA'y'take A'P=AP. 


In the faces X'V'Y' and X'V'Z' construct P'Q' and P'R’ 
respectively, each J_ to V'A'. 


Draw 

Q'R'. 


Since 

AP=A'P', 

Const. 

since 

ZQPA = ZQ'P'A', 

Post. 6 

and since 

ZPAQ = ZP'A'Q', 

§38 

then 

AAPQ is congruent to AA'P'Q'. 

§44 

Hence 

AQ=A'Q' 


and 

PQ = P'Q'. 

§38 

Similarly, 

AR=A’B' and PR = P’R'. 


Now, since APAC is congruent to AP'A'C', 

Proved 

we have 

ACAB = AC'A'B'. 

§38 

Then 

AARQ is congruent to AA'P'Q', 

§40 

and hence 

RQ —R'Q\ 

§38 

Then 

AQPR is congruent to AQ'P'P', 

§47 

and hence 

ARPQ = AR'P'Q'. 

§38 

y 

. dihedral Z VA = dihedral Z F'A'. 

§349 

Similarly, 

dihedral AVB= dihedral ZV'P', 


and 

dihedral AVC = dihedral Z V'C'. 


Hence 

Z V-XYZ=A V-X'Y'Z\ 

§364 

or 

these trihedral A are symmetric. 

§365 


The symmetric angles are shown in the two figures at the right. 


312 


POLYHEDRAL ANGLES 


BOOK VI 


Exercises. Review 

1. Make a list of the numbered propositions in Book VI, 
stating under each the previous propositions employed 
directly in its proof. 

2. Make another list of the numbered propositions, stat¬ 
ing under each the subsequent propositions or corollaries 
in Book VI in which it is used as an authority. 

3. Mention three trihedral angles which may be found 
in a room, and state whether they are equal, symmetric, or 
both equal and symmetric. 

4. By consulting a dictionary find the derivation of 
the words ''dihedral,^^ ^'trihedral,” and polyhedral,’^ and 
explain how these derivations apply to the figures. 

5. If each of two trihedral angles have right angles for 
their face angles, they are both equal and symmetric. 

6. Consider whether or not two trihedral angles are 
equal if two face angles and the included dihedral angle 
of one are equal respectively to two face angles and the 
included dihedral angle of the other, and give the proof. 

7. State a condition under which two polyhedral angles 
of four faces are equal, and prove the equality. 

8. In the trihedral Z.V-ABC, what is the locus of points 
equidistant from the faces VAB and VBCl from the faces 
VAB and VGA ? from the faces VAB and VBC, and also from 
the faces VAB and VCAl To what proposition in plane 
geometry does this correspond ? 

9. If two intersecting planes pass through two parallel 
lines a and b respectively, their line of intersection is also 
parallel to a and h. 

10. A line parallel to each of two intersecting planes is 
parallel to their line of intersection. 


BOOK VII 

POLYHEDRONS, CYLINDERS, AND CONES 

I. Prisms 

368. Polyhedron. A solid bounded by planes is called a 
polyhedron. 

For example, in the second figure of § 367, if we consider that part 
of the Z.V-XYZ cut off by the plane ABC, we have a polyhedron. 

The bounding planes are called the faces of the polyhedron, the 
intersections of the faces are called the edges of the polyhedron, and 
the intersections of the edges are called the vertices of the polyhedron. 

A line joining any two vertices not in the same face is called a 
diagonal of the polyhedron. 

If every plane which cuts a given polyhedron forms a convex polygon, 
the polyhedron is said to be convex. We shall consider only convex 
polyhedrons in this course. 

369. Prism. A polyhedron of which two faces are con¬ 
gruent polygons in parallel planes, and the other faces are 
parallelograms, is called a prism. 

The figure at the right shows a prism. 

The parallel polygons are called the bases 
of the prism, the parallelograms are called 
the lateral faces, and the intersections of the 
lateral faces are called the lateral edges. 

The lateral edges of a prism are equal (§ 76). 

The sum of the areas of the lateral faces 
is called the lateral area of the prism. 

The perpendicular distance between the 
planes of the bases is called the height or altitude of the prism. 

The meaning of the terms congruent and equivalent as applied to 
polyhedrons is evident from plane geometry. 

PS 313 






314 


PRISMS 


BOOK VII 


370. Right Prism. A prism whose lateral edges are per¬ 
pendicular to its bases is called a right prism. 

The first of the figures below shows a right prism. The lateral 
edges of a right prism are equal to the altitude (§ 341). 

371. Oblique Prism. A prism whose lateral edges are 
oblique to its bases is called an oblique prism. 

The second of the figures below shows an oblique prism. 



372. Prisms classified as to Bases. Prisms are said to be 
triangular, quadrangular, and so on, according as their 
bases are triangles, quadrilaterals, and so on. 

Thus, the first figure above shows a quadrangular prism, the second 
shows a triangular prism, and so on. 

373. Right Section. The polygon formed by the intersec¬ 
tions of the lateral faces of a prism with a plane which 
cuts all the lateral edges, produced if necessary, and is 
perpendicular to them, is called a right section. 

The third figure above shows how a right section is formed. 

374. Truncated Prism. The part of a prism included be¬ 
tween the base and a section made by a plane oblique to 
the base is called a truncated prism. 

The fourth of the above figures shows a truncated prism. 













































§§ 370-375 


SECTIONS OF A PRISM 


315 


Proposition 1. Sections of a Prism 

375. Theorem. The sections of a prism made by par¬ 
allel planes cutting all the lateral edges are congruent 
polygons. 



Given the prism PR and the sections AD and A^D' made by 
II planes cutting all the lateral edges. ' ^ 

Prove that AD is congruent to A'D'. 

Proof. Lettering the figure as shown above, we see_that 
AB is II to A'B\ BC is II to^'C', CD is II to C'D\ 


and so on for all the corresponding sides. § 338 

If two II planes are cut by a third plane, the lines 
of intersection are II. 

Then AB=A'B', BC==B'C', CD = C'D', 
and so on for all the corresponding sides. § 76 

The opposite sides of a EJ are equal • • •. 

Also, ZCBA = ZC'B'A\ ZDCB = ZD'C'B\ 

and so on for all the corresponding A. § 343 

AD is congruent to A'D'. § 37 


Since all their corresponding parts are equal, the sections can he 
made to coincide by superposition. 

As a special case of this theorem, all right sections of a prism are 
congruent. 









316 


PRISMS 


BOOK VII 


Proposition 2. Lateral Area 

376. Theorem. The lateral area of a prism is the prod¬ 
uct of a lateral edge and the perimeter of a right section. 



Given FF, a rt. section of the prism AD'; L, the lateral area; 
e, a lateral edge; and p, the perimeter of the rt. section. 

Prove that L — ep. 

Proof. Lettering the figure as shown, we see that 



AA'=BB'=CC'=DD'=EE'=e. 

§369 

Also, 

VWis ± toBB'. 

§316 

Similarly, 

WX is ± to CC', and so on. 



nAB' = BB' -VW^e- VW. 

§243 

Similarly, 

nBC'^CC ■ WX-= e ■ WX, 


and so on for all the lateral faces. 


Now 

L is the sum of these areas. 

§369 


L = e{VW+WX + XY+YZ + ZV). 

Ax.l 

But 

VW+WX + XY+YZ+ ZV= p. 

§7 


II 

Ax. 5 













§376 


LATERAL AREA 


317 


Exercises. Practical Measurements 

1. The lateral area of a right prism is the product of 
the altitude and the perimeter of the base. 

2. The right section of a steel rod 10 ft. long is a square 
whose area is 4.41 sq. in. If the lateral surface is to be 
nickeled, how many square inches are to be covered ? 

3. The right section of an iron rod 8 ft. long is an equi¬ 
lateral triangle whose area is Vs sq. in. If the lateral sur¬ 
face is enameled, how many square inches are covered ? 

4. The lateral surface of an iron bar 6 ft. long is to 
be gilded. If the right section is a square whose area is 
1.69 sq. in., how many square inches are covered ? 

5. A right prism is 2| in. long and its basejs an equi¬ 
lateral triangle whose altitude is 0.866 in. (or I- \/3 in.). Find 
the lateral area. 

6. Find the total area of a right prism twice as long as 
it is thick, and whose base is a square 5.76 sq. in. in area. 

7. What is the total area of a right prism whose altitude 
is 28 in., and whose base is a right triangle of which the 
hypotenuse is 70 in. and one side is 42 in.? 

Find the lateral areas of the right prisms whose altitudes 
{h) and perimeters {p) of bases are as follows: 

8. = 16 in., p = 27 in. 9. A = 2 ft. 9 in., p = 3 ft. 8 in. 

Find the lateral areas of the prisms whose lateral edges 
(e) and perimeters {p) of right sections are as follows: 

10. e = 15 in., p = 28 in. 11. e = 1 ft. 7 in., p = 2 ft. 8 in. 

Find the lateral edges of the prisms whose lateral areas 
(L) and perimeters {p) of right sections are as follows: 

12. L = 169, p = 2. 13. L = 225 sq. ft., p = 9 ft. 


318 


PRISMS 


BOOK VII 


Proposition 3. Congruent Prisms 

377. Theorem. If the three faces which include a tri¬ 
hedral angle of one 'prism are congruent respectively to 
three faces which include a trihedral angle of another, 
and are similarly placed, the prisms are congruent 



Given the prisms AI, with faces AD, AG, AJ congruent 
respectively to faces A'G\ A'J', and similarly placed. 

Prove that AI is congruent to A'I'. 

Proof. Lettering as shown, we see that ABAE = ZB'A'E', 
ZBAF = ZB'A’F', ZEAF = ZE’A'F' (§ 38), and that they 
are arranged in the same order (given). 

.*. trihedral ZA= trihedral ZA\ § 367 

Now show by superposition that face AD can be made to coincide 
with face A'D', face AG with face A'G', and face AJ with face A'J'; 
C lying on C', and D on D'. 

Then show by §§ 52, 314 that the planes of the upper bases coincide. 

Finally show that the prisms coincide and hence are congruent. 

378. Corollary. Two truncated prisms are congruent under 
the conditions given in § 377. 

379. Corollary. Two right prisms which have congruent 
bases and equal altitudes are congruent. 










§§ 377-380 


EQUIVALENT PRISMS 


319 


Proposition 4. Equivalent Prisms 

380. Theorem. An oblique prism is equivalent to a 
right prism whose base is equal to a right section of 
the oblique prism, and whose altitude is equal to a 
lateral edge of the oblique prism. 



Given FI, a rt. section of the oblique prism AD\ and FF, 
a rt. prism whose altitude is equal to a lateral edge of AD\ 

Prove that AD' is equivalent to FI'. 

Proof. Lettering the figure as shown, we see that if we 
take from the equal lateral edges of AD' and FI' (§ 370) the 
lateral edges of FD', which are common to both, we have 
AF=A'F',BG = B'G',CH=C'H'r ''- Ax. 2 

Also, base FI is congruent to base F'l'. § 369 

Now prove the truncated prisms AI and A'F congruent (§ 378). 


Then 

AI-\-FD'=A'l'->rFD'. 

Ax.l 

Now 

AI-\-FD'=^AD', 


and 

A'l'-\-FD'==Fl'. 

Ax. 10 


. ad' is equivalent to FI'. 

Ax. 5 









320 


PRISMS 


BOOK VII 


381. Parallelepiped. A prism whose bases are parallelo¬ 
grams is called a parallelepiped. 

382. Right Parallelepiped. A parallelepiped with edges 
perpendicular to the bases is called a right parallelepiped. 

383. Rectangular Parallelepiped. A right parallelepiped 
whose bases are rectangles is called a rectangular paral¬ 
lelepiped. 

The first figure below shows a rectangular parallelepiped. The four 
lateral faces of this solid are also rectangles. 



The terms 'parallelepiped, right parallelepiped, and rectangular 
parallelepiped are not commonly used outside the school, the last being 
generally called a rectangular solid or a square solid. Some teachers 
employ the term cuboid, but it has not come into practical use. 

Among the common illustrations of rectangular solids are boxes, 
bricks, rooms, and the like. 

The oblique parallelepiped illustrated in the third figure above is 
rarely found in practice, and no special definition is necessary. 

384. Cube. A parallelepiped whose six faces are all 
squares is called a cube. 

The second figure above shows a cube. It might properly be asked 
how we know that such a figure is possible. We may, for example, 
speak of a seven-edged polyhedron, but such a figure does not exist. 
In elementary work, however, it is not expected that attention will be 
given to such details. 





















§§ 381-385 


PARALLELEPIPEDS 


321 


Proposition 5. Opposite Faces 

385. Theorem. The opposite faces of a parallelepiped 
. are congruent and parallel. 



Given the parallelepiped AC. 

Prove that the faces AB\ DC' and the faces AD', BC' are 
respectively congruent and II. 

What previous proposition can be used to prove that AB' is II to DC' ? 
Can it be proved by superposition that AB' is congruent to DC' ? The 
proof is left for the student. 


Exercises. Parallelepipeds 

1. The diagonals of a parallelepiped bisect each other. 

2. The lateral faces of a right parallelepiped are rec¬ 
tangles, and its four diagonals are equal. 

3. Compute the lengths of the diagonals of a rectangular 
parallelepiped whose edges meeting at any vertex are a, h, c. 

4. Every section of a parallelepiped made by a plane 
parallel to the lateral edges is a parallelogram. 

5. Investigate Ex. 4 for any prism whatever. 

6. Given a rectangular parallelepiped, lettered as in the 
figure of § 385, with AB = 8, BC = 6, and CC' = find the 
length of the diagonal AC'. 








822 


PRISMS 


BOOK VII 


Proposition 6. Diagonal Plane of a Parallelepiped 

386. Theorem. The plane passed through diagonally 
opposite edges of a parallelepiped divides the parallele¬ 
piped into two equivalent triangular prisms. 



Given the plane m passed through the opposite edges AA^ and 
CC of the parallelepiped AC. 

Prove that AC' is divided into two equivalent triangular 


prisms ABC-B' and CDA-D'. 

Proof. Let WXYZ be a right section of AC'. § 373 

Then faces AB' and DC' are congruent and II, 

and faces and BC' are congruent and II. § 385 

Hence in the section WXYZ, 

WX is II to ZY, and XY is II to WZ. § 338 

Hence WXYZ is a O, § 72 

and AWXY is congruent to AYZW. § 77 

Now prism ABC-B' is equivalent to a rt. prism with 
base WXY and altitude A A', and prism CDA-D' is equivalent 
to a rt. prism with base YZW and altitude AA'. § 380 

The rest of the proof is left for the student. 















§386 


PARALLELEPIPEDS 


323 


Exercises. Practical Measurements 

1. Given that the three plane angles at one of the 
vertices of a parallelepiped are 80°, 70°, and 75° respectively, 
find all the other angles in all the faces. 

2. The three edges of the trihedral angle at one of the 
vertices of a rectangular parallelepiped are 10 in., 12 in., 
and 14 in. respectively. Find the area of the total surface 
of the parallelepiped. 

3. The three face angles at one vertex of a parallelepiped 
are each 60°, and the three edges of the trihedral angle at 
that vertex are 6 in., 4 in., and 2 in. respectively. Find the 
area of the total surface to the nearest 0.01 sq. in. 

4. In a rectangular parallelepiped the square of any 
diagonal is equivalent to the sum of the squares of any 
three edges which meet at one vertex. 

5. In a box 6 in. deep and 12 in. wide, a wire 2 ft. long 
reaches from one corner to the diagonally opposite corner. 
Find the length of the box to the nearest 0.01 in. 

6. The height of a rectangular parallelepiped is 22 in. 
and the length of the diagonal of the base is 30 in. Find the 
length of the diagonal of the parallelepiped. 

7. The total area of the six faces of a cube is 108 sq. in. 
Find the length of the diagonal of the cube. 

8. The diagonal of the face of a cube is V6. Find the 
diagonal of the cube. 

9. The diagonal of a cube is 5 J Vs. Find the diagonal 
of a face of the cube. 

10. A water tank is 4 ft. long, 3 ft. wide, and 2 ft. deep. 
How many square feet of zinc will be required to line the 
four sides and the base, allowing 2 sq. ft. for overlapping 
and for turning the top edge ? 


324 


PRISMS 


BOOK VII 


11. A square sheet of galvanized iron 8 ft. on a side leans 
against a wall and is inclined at an angle of 60° to the 
horizontal. What area of ground does the sheet protect 
from rain falling vertically ? 

12. Through a point P on the sheet of iron in Ex. 11, what 
line in the inclined plane will make the largest angle with 
the horizontal ? Give the reason for your answer, and state 
the number of degrees in this angle. 

13. A rectangular solid has for its lower base the □ ABCD 
and for its upper base the \Z\A'B'C'D\ lettered in the cor¬ 
responding way. What plane passing through the diagonal 
DB' is II to AB, and what angle does the plane make with 
the lower base ? 

14. The length of the diagonal of a rectangular solid 
is 17 in. and the area of the total surface is 552 sq. in. Find 
the sum of the three dimensions. 

15. If the length, width, and height of a room are a, h, 
and c respectively, what is the total area of the four walls ? 
of the walls, floor, and ceiling ? 

16. The outside dimensions of a closed wooden box are 
8 in., 10 in., and 12 in., and the area of the total inside sur¬ 
face is 376 sq. in. Find the thickness of the wood used in 
making the box. 

17. The area of the total surface of a rectangular block 
is 1332 sq. in. and the dimensions are proportional to 4, 5, 
and 6. Find the dimensions. 

18. The lower base of a cube which is \/2 on an edge 
is ABCD, and the upper base \s>A'B'C'D\ lettered in the 
corresponding way. If a plane passes through A', C', and 
B, what is the area of the AA'BC'^l 

19. The area of AA'BC' in Ex. 18 is what part of the 
area of the diagonal plane AB'C'Dl 


325 


§§387-392 VOLUME OF A PARALLELEPIPED 

387. Unit of Volume. In measuring volumes, a cube whose 
edges are all equal to the unit of length is taken as the unit 
of volume. 

Thus, if we are measuring the contents of a box of which the dimen¬ 
sions are given in feet, we take 1 cu. ft. as the unit of volume. 

388. Volume. The number of units of volume contained 
by a solid is called its volume. 

389. Equivalent Solids. Two solids which have equal vol¬ 
umes are said to be equivalent. 

390. Dimensions. The lengths of the three edges of a rec¬ 
tangular parallelepiped which meet at a common vertex are 
called its dimensions, 

391. Volume of a Rectangular Parallelepiped. Assuming 
that the three edges are commensurable, suppose that Z, 
the length, contains 4 units; that w, the 
width, contains 2 units; and that h, the 
height, contains 5 units. Then y, the vol¬ 
ume, contains 4 X 2 X 5, or 40, cubic units. 

In general, if there are I units of length, 
w units of width, and h units of height, then 

V=lwh\ 

that is, the volume of a rectangular parallelepiped is the 
product of its three dimensions. 

For on each square unit of base there is one cubic unit for every 
unit of height. Then, since there are Iw square units of base (§ 241), 
there are Iw cubic units for every unit of height. Hence for h units of 
height, there are Iwh cubic units of volume. 

The incommensurable case is considered in § 517. 

392. Corollary. The volume of a rectangular parallelepiped 
is the product of its base and altitude. 

For, if B is the area of the base, B = Iw, and hence V = Bh. 










/ 

h 




* 











< - 1 - 

















320 


PRISMS 


BOOK VII 


Proposition 7. Volume of a Parallelepiped 

393. Theorem. The volume of any parallelepiped is 
the product of the base and the altitude. 



Given P, a parallelepiped with no two faces _L to each other; 
F, the volume; B, the area of the base; and h, the altitude. 

Prove that V=Bh. 

Proof. Produce the edge CD and the edges II to CD, and 
cut them perpendicularly by two II planes whose distance 
apart, EF, is equal to CD. We then have the oblique paral¬ 
lelepiped Q whose base is a □. 

Produce EG and the edges II to EG, and cut them perpen¬ 
dicularly by two II planes whose distance apart, HI, is equal 
to EG. We then have the rectangular parallelepiped R. 


Since P=Q, and Q = R, § 380 

then P = R. Ax. 5 

Also, P, Q, and R have the common altitude h. § 341 

Let B' be the area of the base of Q, and B" that of R. 

Since B = B\ and B' = B", § 245 

then B = B\ Ax. 5 

Now the volume of R is B''h. § 392 

.\V=Bh. Ax. 5 










§393 


VOLUMES 


327 


Exercises. Parallelepipeds 

1. Two rectangular parallelepipeds with congruent bases 
are to each other as their altitudes. 

In all such cases the words '' the volumes of ’ ’ are understood. 

2. Two rectangular parallelepipeds with equal altitudes 
are to each other as their bases. 

3. Two rectangular parallelepipeds with one dimension 
in common are to each other as the products of their other 
two dimensions. 

4. Two rectangular parallelepipeds with two dimensions 
in common are to each other as their third dimensions. 

5. Two rectangular parallelepipeds are to each other as 
the products of their three dimensions. 

6. The volume of any parallelepiped is equal to that of 
a rectangular parallelepiped of equivalent base and equal 
altitude. 

7. Find the ratio of two rectangular parallelepipeds 
which have the dimensions a, 6, c, and 2 a, 3 6, and 4 c 
respectively. 

8. Find the ratio of two rectangular parallelepipeds 
both of whose altitudes are h inches, and whose bases are a 
inches by 2 6 inches, and 2 6 inches by 3 a inches respectively. 

9. Find the volume of a rectangular parallelepiped I feet 
long, w inches wide, and h yards high, expressing the result 
as cubic inches; as cubic feet; as cubic yards. 

10. Find the volume of a rectangular parallelepiped 
whose base is B square feet and whose altitude is h inches, 
expressing the result as cubic inches; as cubic feet. 

11. The volume of a parallelepiped is a^ cubic inches and 
the area of the base is a square feet. Express the altitude 
in inches; in feet. 


328 


PRISMS 


BOOK VII 


Exercises. Practical Measurements 

1. The volume of a rectangular parallelepiped with a 
square base is 84 cu. in. and the altitude is 6 in. Find the 
dimensions. 

2. A rectangular tank full of water is 9 ft. long and 
5 ft. 6 in. wide. How many cubic feet of water must be 
drawn off in order that the surface may be lowered 1 ft? 

In all such cases the measurements are supposed to be made on the 
inside unless the contrary is stated. 

3. What dimensions should be allowed for a rectan¬ 
gular container which shall hold just 1 gal. (231 cu. in.) if 
each dimension must be a whole number of inches ? 

4. The inside dimensions of a covered box, made of 
steel I in. thick and weighing 490 lb. per cubic foot, are 
16 in., 9 in., and 4 in. Find the total weight of the box. 

5. A steel bar 6 ft. long is 2 in. wide and if in. thick. 
At 490 lb. per cubic foot, how much does it weigh ? 

6. If 3 cu. in. of gold beaten into gold leaf will cover 
75,000 sq. in. of surface, what is the thickness of the leaf ? 

7. The sum of the squares of the four diagonals of a 
parallelepiped is equal to the sum of the squares of the 
twelve edges. 

8. The volume of a cube is 216 cu. in. Find to the 
nearest 0.01 in. the length of the diagonal. 

9. Find a formula for the total surface of a cube in 
terms of the diagonal of the cube. 

10. If the total surface of one cube is n times that of 
another cube, the volume of the first is how many times the 
volume of the second ? 

11. Find the weight of a wooden beam 8 in. by 10 in., 
18 ft. long, and weighing 54 lb. per cubic foot. 


§394 


TRIANGULAR PRISM 


329 


Proposition 8. Triangular Prism 

394. Theorem. The volume of a triangular prism is 
the product of the base and the altitude. 



Given PQR-Q'^ a triangular prism; F, the volume; By the 
area of the base; and /z, the altitude. 

Prove that V = Bh. 

Proof. Since, in any plane, we can construct a line li to 
a given line, we can construct PS II to QR, RS II to QP, 
SS' II to PP'y P'S' II to PS, and R'S' II to RS, thus forming 
the parallelepiped PQRS-Q'. 

Then pqr-Q'=1 PQRS-Q'. § 386 

But PQRS-Q'=PQRS • h, § 393 

and PQRS=2B. §77 

Hence PQRS-Q' =2B>h, 

and PQR-Q' = l{2B'h), Ax. 5 

PQR-Q'=Bh. 

Substituting V for PQR-Q', we have 
V=Bh. 


or 















330 


PRISMS 


BOOK VII 


Proposition 9. Volume of a Prism 

395. Theorem. The volume of any prism is the product 
of the base and the altitude. 



Given PR\ a prism; F, the volume; 15, the area of the base ; 
and hy the altitude. 

Prove that V = Bh. 

Proof. From the vertex R in the lower base PQR ST draw 
the diagonals RP and RT. 

From the vertex R' in the upper base P'Q'R'S'T' draw 
the diagonals R'P' and R'T'. 

Taken together, the triangular prisms thus determined 
form the given prism. 

Similarly, taken together, the respective bases of the 
triangular prisms form the bases of the given prism. 

Now the volume of each triangular prism is the product 
of the base and the altitude h. § 394 

Hence the total volume of the prism PR' is the sum of 
the bases of the triangular prisms multiplied by the com¬ 
mon altitude h. Ax. 1 

That is. 


V=Bh. 


















§395 


REVIEW EXERCISES 


331 


Exercises. Review 

1. Prisms with equivalent bases and equal altitudes are 
equivalent. 

2. Prisms with equivalent bases are to each other as 
their altitudes; prisms with equal altitudes are to each 
other as their bases. 

3. The volume of a triangular prism is equal to half the 
product of a lateral face and the distance of this face from 
the opposite edge. 

4. The edge of a cube is e. Write a formula for the sum 
of the edges; for the area of the total surface; for the 
xliagonal of the cube. 

5. The edge of a cube is e. Find the volume of a cube 
that has an edge twice as long. Find the edge of a cube 
of twice the volume. 

6. The altitude of a prism is 7 in., and the base is an 
equilateral triangle which is 7 in. on a side. Find the vol¬ 
ume of the prism. 

7. The number of square millimeters in the area of the 
total surface of a certain cube is equal to the number of 
cubic millimeters in the volume of the cube. Find the 
length of 'each edge. 

8. In Ex. 7 find the length of the diagonal of the cube. 

9. The number of square millimeters in the area of the 
total surface of a right triangular prism with an equilateral 
base is equal to the number of cubic millimeters in the 
volume, and the altitude of the prism is equal to the side 
of the base. Find the altitude. 

10. If the base of a right prism is a regular polygon 
of apothem a, and the area of the lateral surface is L, 
the volume V is given by the formula aL. 


332 


PRISMS 


BOOK VII 


Exercises. Practical Measurements 

1. If the length of a rectangular parallelepiped is 22 in., 
the width 8 in., and the height 6 in., what is the area of the 
total surface ? 

2. Find the volume of a triangular prism whose height 
is 14 in. and whose base has the sides 10 in., 8 in., and 6 in. 

3. Find the volume of a prism whose height is 12 ft. and 
whose base is an equilateral triangle 10 in. on a side. 

4. The base of a right prism is a rhombus of which one 

side is 10 in., and the shorter diagonal 12 in. The height of 
the prism is 15 in. Find the area of the total surface and 
the volume of the prism. » 

5. An open tank 8 ft. long and 51' ft. wide holds 264 cu. ft. 
of water. How many square feet of sheet lead will it take 
to line the sides and bottom ? 

6. How much sheet lead will be required to line an open 
tank which is 5 ft. long, 3 ft. 6 in. wide, and contains 
105 cu. ft.? 

7. The diagonal of one of the faces of a cube is V? in. 
Find the volume of the cube. 

8. The three dimensions of a rectangular parallelepiped 
are a, h, c. Find in terms of a, h, and c the volume, the area 
of the total surface, and the length of the diagonal. 

9. If the height of a prism is 5 in., and the base is a 
regular hexagon 1 in. on a side, what is the volume ? 

10. An open cistern is made of iron \ in. thick, and the 
inside dimensions are as follows: length, 6 ft.; width, 4 ft.; 
depth, 3ft. What will the cistern weigh when empty? 
when full of water? 

A cubic foot of water weighs 62^ lb. The specific gravity of iron 
is 7.2; that is, iron is 7.2 times as heavy as water. 


§§ 396-398 PYRAMIDS 333 

f 

II. Pyramids 

396. Pyramid. A polyhedron of which one face, called the 
base, is any polygon and the other faces are triangles with 
a common vertex is called a pyramid. 

The triangular faces with the common vertex are called the lateral 
faces, their intersections are called the lateral edges, and their common 
vertex is called the vertex of the pyramid. In our work we shall con¬ 
sider only pyramids whose bases are convex polygons. 

The spm of the areas of the lateral faces is called the lateral area 
of the pyramid. The perpendicular distance from the vertex to the 
plane of the base is called the height or altitude of the pyramid. 



397. Pyramids classified as to Bases. Pyramids are said to 
be triangular, quadrangular, and so on, according as their 
bases are triangles, quadrilaterals, and so on. 

A triangular pyramid is also called a tetrahedron. 

398. Regular Pyramid. If the base of a pyramid is a 
regular polygon whose center coincides with the foot of 
the perpendicular from the vertex to the base, the pyramid 
is called a regular pyramid. 

The altitude of the triangle which forms one of the lateral faces of 
a regular pyramid is called the slant height of the pyramid. 

The figures above show different types of pyramids, the two at the 
right being regular pyramids. 












334 


PYRAMIDS 


BOOK VII 


399. Properties of Regular Pyramids. The proofs of the 
following obvious properties of regular pyramids depend 
upon the theorems indicated: 

1. The lateral edges of a regular pyramid are 
equal (§ 326). 

For they meet the base at equal distances from the 
center (§ 398). 

2. The lateral faces of a regular pyramid are 
congruent isosceles triangles (§47). 

3. The slant height of a regular pyramid is the same for 
all the lateral faces (§ 326). 

400. Frustum of a Pyramid. The portion of a pyramid 
included between the base and a section parallel to the 
base is called Si frustum of a pyramid. 

The figure at the right shows a frustum 
of a regular pyramid. 

A more general term, including a frustum 
as a special case, is truncated pyramid, which 
is applied to the portion of a pyramid in¬ 
cluded between the base and any section 
whatever made by a plane that cuts all the 
lateral edges. This term, however, is little 
used at the present time. 

The base of the pyramid and the parallel section are called the bases 
of the frustum. 

The perpendicular distance between the bases is called the height 
or altitude of the frustum. The altitude is 
represented by h in the figure here shown. 

The portions of the lateral faces of a pyra¬ 
mid that lie between the bases of a frustum 
are called the lateral faces of the frustum, 
and the sum of their areas is called the lateral 
area of the frustum. 

The altitude of one of the lateral faces of a frustum of a regular 
pyramid is called the slant height of the frustum. The slant height is 
represented by I in the above figure. 

















§§ 399-402 


LATERAL AREA 


335 


Proposition 10. Lateral Area of a Pyramid 

401. Theorem. The lateral area of a regular pyramid 
is half the product of the slant height and the perimeter 
of the base. 



Given V-ABCDE^ a regular pyramid; L, the lateral area; 
/, the slant height; and p, the perimeter of the base. 


Prove that L — \lp. 

Proof. The lateral faces are congruent A. § 399,2 
Now the area of each face is JZ times the base, § 244 
and the sum of the bases of the A is p. § 7 

Then the sum of the areas of the A is ^ Ip. Ax. 1 
. *. L = \ Ip. Ax. 5 


402. Corollary. The lateral area of a frustum of a regular 
pyramid is half the product of the slant 
height of the frustum and the sum of the 
perimeters of the bases. 

How is the area of a trapezoid found (§ 247)? 

Are the faces congruent trapezoids ? What is the 
sum of their lower bases ? of their upper bases ? 

What is the sum of their areas? Write the for¬ 
mula and give the proof in full. 












336 


PYRAMIDS 


BOOK VII 


Proposition 11. Section Parallel to Base 

403. Theorem. If a 'pyramid is cut hy a plane parallel 
to the basey 

1. The lateral edges and the altitude are divided pro¬ 
portionally, 

2. The section is a polygon similar to the base. 

3. The area of the section is to the area of the base as 
the square of the distance of the plane from the vertex 
is to the square of the altitude of the pyramid. 



Given the pyramid V-ABCDE cut by tti, a plane li to the 
base and intersecting the altitude FO in O'y emd A'B’C'D'E'y 
the section thus formed. 

Prove that ^ = that A’B'C'D'E' is 

similar to ABODE; and that ^^ ■ 

ABODE yo 

1. Use §§ 338 and 201 to prove the first proportion. 

2. Prove AVA'B' similar to AVAB, and so on. Then prove the neces¬ 
sary conditions (§ 205) under which two polygons are similar. 

3. Prove that A'B'CD'E ': ABCDE=A ^'^: AB'^ =VO''^ :VO'\ 













§§ 403 , 404 


SECTION PARALLEL TO BASE 


337 


404. Corollary. If twa pyramids have equal altitudes and 
equivalent bases, sections made by planes parallel to the bases 
at equal distances from the vertices are equivalent 


A'B'C'D'E' VO'‘^ 
ABODE ~ FO" ' 

X'Y'Z' 

XYZ 

VO' ^ WP' 
VO WP’ 

. VO'^ ^ WP'‘" 

Fo " ~ Wp' 

Then 



A'B'C'D'E' __ X'Y'Z' 
ABODE ~ XYZ 


and hence 


A'B'C'D'E'= X'Y'Z'. 


§ 198,2 


Exercises. Review 


1. The base of a pyramid is an equilateral triangle 2 in. 
on a side. Find the area of a section parallel to the base 
and halfway between the vertex and the base. 

2. A section of a pyramid parallel to the base is equal to 
half the base. If the altitude of the pyramid is 10 in., how 
far is the section from the base ? 

3. Solve Ex. 2 when the section is one nth. of the base 
and the altitude of the pyramid is h feet. 

4. The perimeter of the base of a regular pyramid is 
40 mm. and the lateral area is 320 sq. mm. Find the slant 
height of the pyramid. 

5. The top of a certain obelisk is a regular pyramid with 
a square base 225 sq. in. in area and an altitude of 30 in. 
Find to the nearest 0.1 in. the slant height of the pyramid, 
and then find the lateral area. 










338 


PYRAMIDS 


BOOK VII 


Proposition 12. Triangular Pyramids 

405. Theorem. Two triangular pyramids with equiv¬ 
alent bases and equal altitudes are equivalent. 



Given the triangular pyramids P and P' with the equivalent 
bases ABC and and the equal altitudes h. 


Prove that P = P'. 

Proof. Place the bases in the plane m, divide h into any 
number of equal parts, as a, and through the points of divi¬ 
sion of h let the planes n, p, q, • • • pass II to m, as shown. 

Let the prisms y, z, - • with lateral edges II to VA have 
DEF, GHI, • • • for their upper bases, and let the prisms 
x', y\ ' with lateral edges II to V'A' have A'B'C\ D'E'F\ 
G'H'I',' ■ ‘ for their lower bases. 

In the above figure there are two prisms in one case and three in 
the other, but the proof may be applied to any number. 


Since 

and 

we have 
Similarly, 


DEF = D'E'F' 
the altitudes are equal, 

y = y'- 

z — z. 


§404 

Const. 

§395 























§405 


TRIANGULAR PYRAMIDS 


389 


Hence x' y' z' — {y-\-z) = x'. Ax. 2 

Then by substituting P\ which is less than x'-\-y'-\-z\ 
for cc'H- 2/'+ and by substituting P, which is greater than 
y-\-z, for y-\-z,v^e have 

P'-P<x\ 

That is, the difference between the pyramids is less than 
x\ which is the difference between the sets of prisms. 

Now by increasing indefinitely the number of parts into 
which h is divided, and consequently decreasing a indefi¬ 
nitely, x' can be made as small as we please. 

Hence whatever difference we assume to exist between 
the pyramids, x' can be made smaller than that difference. 

But this is impossible, since we have shown that x' is 
greater than the difference, if any exists. 

Hence it leads to an impossibility to suppose that 
P'>Py or that P'<P. 

.*. P = P\ 

Exercises. Review 

1. The slant height of a regular pyramid is 12 in., and the 
base is an equilateral triangle whose altitude is 4V3in. 
Find the lateral area. 

2. The slant height of a regular triangular pyramid is* 
equal to the altitude of the base, and the area of the base 
is \/5 sq. ft. Find the area of the total surface. 

3. If one pyramid has for its base a right triangle with 
hypotenuse 10 and shortest side 6, and another pyramid of 
equal altitud e has for its base an equilateral triangle which 
is 4 V2 Vs on a side, the pyramids are equivalent. 

4. The base of one of two equivalent pyramids 6 in. high 
is 8 sq. in. in area, and that of the other is an equilateral 
triangle. Find the lateral area of the second pyramid. 



340 


PYRAMIDS 


BOOK VII 


Proposition 13. Volume of a Triangular Pyramid 

406. Theorem. The volume of a triangular pyramid is 
one third the product of the base and the altitude. 



Given P-QRSy a triangular pyramid; F, the volume ; the 
area of the base; and /i, the altitude. 

Prove that V=i Bh. 

Proof. On QRS as base let there stand the prism QRS-XPY, 
and let the plane XPS pass through XP and S. 

Then the prism is composed of the three triangular pyra¬ 
mids P-QRS, P-SYX, and P-QSX. 


Now 

P-SYX and P-QSX have the same altitude, 

§396 

and 

base SYX = base QSX. 

§77 


.*. P-SYX = P-QSX. 

§405 

But 

P-SYX is the same as S-XPY. 


Now 

S-XPY has the same altitude as P-QRS, 

§341 

and 

base XPY= base QRS. 

§369 


.'. P-SYX = P-QRS. 

§405 

Then 

P-QRS = P-SYX = P-QSX, 

Ax. 5 

and hence P-QRS = J of QRS-XPY. 




§394 












§§ 406,407 


VOLUME 


341 


Proposition 14. Volume of any Pyramid 

407. Theorem. The volume of any pyramid is one third 
the product of the base and the altitude. 



Given P-QRSTU^ a pyramid; F, the volume; By the area of 
the base; and /z, the altitude. 

Prove that V=\Bh. 

Proof. From any vertex of the base draw diagonals to 
the other vertices. In the above figure, let these diagonals 
be TQ and TR. 

Then the planes determined by PT and the diagonals 
TQ, TR divide the given pyramid into three triangular 
pyramids, each of which has the altitude h. 

The volume of each of these triangular pyramids is 


^ h times the area of the base. § 406 

Hence the volume of P-QRSTU, the sum of the triangular 
pyramids, \s>\h times the sum of the bases. Ax. 1 

But the sum of the bases is B. Ax. 10 

,\V=lBh. Ax. 5 


The proof is evidently the same whatever the number of the tri¬ 
angular pyramids into which the given pyramid is divided. 














342 


PYRAMIDS 


BOOK VII 


Exercises. Properties of Pyramids 

1. The volume of a triangular pyramid is equal to one 
third the volume of a triangular prism of the same base and 
altitude. 

2. The volumes of two pyramids are to each other as the 
products of their bases and altitudes. 

3. Pyramids with equivalent bases are to each other as 
their altitudes. 

4. Pyramids with equal altitudes are to each other as 
their bases. 

5. Pyramids with equivalent bases and equal altitudes 
are equivalent. 

6. In the tetrahedron (§ 397) V-ABC, the midpoints of 
VA, VC, BA, BC are vertices of a parallelogram. 

7. The lines joining the midpoints of the opposite edges 
of a tetrahedron meet in a point. 

In the figure described in Ex. 6 the opposite edges are VA and BC; 
VB and AC; VC and AB. 

8. The plane which passes through an edge of a tetra¬ 
hedron and the midpoint of the opposite edge divides the 
tetrahedron into two equivalent tetrahedrons. 

9. Given that a regular triangular pyramid has all its 
four faces congruent, and that its volume is known, show 
how to find the area of the total surface. 

10. Show how to find the volume of any polyhedron by 
dividing the polyhedron into pyramids. 

11. Given the edge a of the base and the area T of the 
total surface of a regular pyramid with a square base, find 
the height h in terms of a and T. 

12. In Ex. 11 find the volume V in terms of a and T. 


§407 


EXERCISES 


343 


Exercises. Measuring the Pyramid 

1. What is the lateral area of a regular pyramid whose 
slant height is 34 in., and the perimeter of the base 57 in.? 

2. Find the volume of a pyramid with an altitude of 
7 in. and a base 9 sq. in. in area. 

3. The base of a regular pyramid is an octagon 3 m. on 
a side and the slant height is 5 m. Find the lateral area of 
the pyramid. 

4. Find the volume of a pyramid with an altitude of 
6.75 m. and a square base whose diagonal is 3\/2m. 

5. The volume of a regular pyramid with a square base 
is 912 cu. ft. and the altitude is 19 ft. Find the lateral area. 

6. The volume of a regular pyramid with a hexagonal 
base is 249.4 cu. m., and the altitude is 8 m. Find the length 
of each side of the base. 

7. The base of a pyramid is a triangle with sides which 
are 6 in., 8 in., and 10 in., and the volume is 240 cu. in. Find 
the height of the pyramid. 

8. A pyramid 12 in. high has a base which is an equi¬ 
lateral triangle 10 in. on a side. Find the volume. 

9. Find the volume of a regular pyramid with a lateral 
edge of 100 ft. and a square base whose side is 40 ft. 

10. Find the volume of a regular pyramid whose slant 
height is 12 ft. and whose base is an equilateral triangle 
inscribed in a circle 10 ft. in radius. 

11. The eight edges of a regular pyramid with a square 
base are equal and the area of the total surface is T. Find 
the edge. 

12. Given the height h and the area Toi the total surface, 
find the base edge a of a regular quadrangular pyramid. 


344 


PYRAMIDS 


BOOK VII 


Proposition 15. Volume of a Frustum 

408. Theorem. The volume of a frustum of a pyramid 
is one third the product of the altitude by the sum of the 
two bases and the mean proportional between them. 



Given F, a frustum; F, the volume ; F, B\ the areas of the 
bases; and h, the altitude. 

Prove that v^\h{B + B' + \/bB'). 

Proof. Let P be the volume of the pyramid from which F 
is cut, and let P' and h' be the volume and altitude respec¬ 
tively of the small pyramid remaining after F is removed. 


Then V= P- P' = \b (h + h')-^B'h\ §407 

Now -jB:^fB'=h-\-h':h\ §403,3; Ax.6 

Solving, h'= h M iFBpJJ') _ 

^b-Vb' b-b’ 


Then V = i \Bh + (B - 

L B — B . 

Simplifying, V=\h{B + B'+y/m'). 


Ax. 5 


This is a subsidiary proposition in mensuration and may be omitted, 
together with pages 345 and 346, without disturbing the sequence. 
















§408 


FRUSTUM OF A PYRAMID 


345 


Exercises. Frustum of a Pyramid 

1. A piece of marble is in the form of a frustum of a 
regular pyramid with a square base. The frustum is 8 ft. 
high and the sides of the bases are 3 ft. and 2 ft. respec¬ 
tively. Taking the weight of 1 cu. ft. of marble as 165 lb., 
find the weight of the piece. 

2. The slant height of a frustum of a regular pyramid is 
10 ft. and the sides of the square bases are 3 ft. and 2 ft. 
respectively. Find the area of the total surface. 

3. How much earth was removed in an excavation which 
is 6 ft. deep, 40 ft. square at the top, and 36 ft. square at 
the bottom ? 

4. A pile of broken stone is in the form of a frustum of 
a pyramid. The lower base is a rectangle 75 ft. long and 
9 ft. wide, the upper base is 50 ft. by 6 ft., and the height 
of the frustum is 6 ft. If the broken stone is spread over 
a road 30 ft. wide to a depth of 3 in., what length of road 
will it cover ? 

5. A pyramid 4 in. high with a base whose area is 16 sq. in. 
is cut by a plane parallel to the base and 2 in. from the 
vertex. Find the volume of the frustum. 

6. A pyramid 6 in. high with a base whose area is 
324 sq. in. is cut by a plane parallel to the base and 2 in. 
from the vertex. Find the volume of the frustum. 

7. The lower base of a frustum of a pyramid is a square 
8 in. on a side. The side of the upper base is half that of the 
lower base, and the altitude of the frustum is the same as 
the side of the upper base. Find the volume of the frustum. 

8. Consider the formula V=\h{B B' ^BB') of §408 
when 0. Discuss the meaning of the result. Also dis¬ 
cuss the case in which B = B\ 


846 


PYRAMIDS 


BOOK VII 


Exercises. Review 

1. The lower base of a frustum of a pyramid is a square 
6 in. on a side. The area of the upper base is half that of 
the lower base, and the altitude of the frustum is 4 in. Find 
to the nearest 0.01 cu. in. the volume of the frustum. 

2. A pyramid has six edges, each 2 in. long. Find to the 
nearest 0.01 cu. in. the volume of the pyramid. 

3. A regular pyramid 8 in. high has a triangular base, 
and the volume of the pyramid is 16 V2 cu. in. Find to the 
nearest 0.01 in. the length of a side of the base. 

4. In Ex. 3 find the area of the total surface. 

5. The base of a regular pyramid is a square I feet on a 
side, and the slant height is s feet. Find the area of the 
total surface. 

6. The lower base of a frustum of a pyramid is a quad¬ 
rilateral whose sides are 5 in., 6 in., 7 in., and 9 in., respec¬ 
tively, and the corresponding sides of the upper base are 
3 in., X inches, y inches, and z inches. Find y, and z. 

7. A schoolroom 30 ft. long, 24 ft. wide, and 14 ft. high 
is ventilated by an electric fan which discharges every 
20 min. a volume of air equal to the volume of the room. 
Find the amount of air discharged per minute. 

8. An oblique pyramid with a square base 10 in. on a 
side is cut by a plane parallel to the base so that the alti¬ 
tude, measured from the vertex, is divided in the ratio 3: 2. 
Find the area of the section. 

9. A frustum of a square pyramid is 18 ft. high and the 
sides of the bases are 1 ft. and 3 ft. respectively. If the 
frustum is divided into three parts by planes passed par¬ 
allel to the bases and dividing the altitude equally, what 
is the volume of each part ? 


§§ 409, 410 


GENERAL POLYHEDRONS 


347 


III. General Polyhedrons 

409. Polyhedrons classified as to Faces. A polyhedron of 
four faces is called a tetrahedron ; one of six faces, a hexa¬ 
hedron; one of eight faces, an octahedron; one of twelve 
faces, a dodecahedron; one of twenty faces, an icosahedron. 

410. Regular Polyhedron. A polyhedron whose faces are 
congruent regular polygons, and whose polyhedral angles 
are equal, is called a regular polyhedron. 



It is proved in § 413 that there cannot be more than five regular 
polyhedrons, and as a matter of fact there are just five. The five 
regular polyhedrons are shown in the above illustration in the order in 
which the polyhedrons are mentioned in § 409. 

These regular solids occupied the attention of Pythagoras and his 
followers (about 550 B.C.). They were also studied so extensively in the 
school of Plato (about 375 B.c.) that they are often known as the Platonic 
Bodies. The early Greek writers connected them in a fanciful way with 
various phenomena of nature. For example, they assigned the tetra¬ 
hedron to fire, the hexahedron to earth, the octahedron to air, the 
icosahedron to water, and the dodecahedron, apparently the last one 
discovered, to the universe. 

Some of the regular polyhedrons are met in the study of 
crystals. Thus, the cube is found in salt crystals and the 
regular octahedron in certain compounds of copper. 

Pages 347-350, while important in the study of crystals, may be 
omitted without affecting the sequence of propositions. 















348 


GENERAL POLYHEDRONS 


BOOK VII 


411. Models of Regular Polyhedrons. In a work printed in 
1525 a great artist, Albrecht Diirer, showed how to make 
paper models of the five regular polyhedrons. His descrip¬ 
tion suggested drawing on stiff paper the diagrams shown 
below, and then cutting along the full lines and pasting 
strips of thin paper on the edges as indicated. By folding 
on the dotted lines and keeping the edges together by the 
pasted strips of paper, the models can be easily made. 



Tetrahedron 


Hexahedron 





412. Relation of Parts of a Polyhedron. If the number of 
edges of a polyhedron is represented by e, the number of ver¬ 
tices by V, and the number of faces by/, then e + 2 = v -f/. 
This remarkable relation was possibly known to the Greek 
mathematician Archimedes (about 250 B.C.), but was first 
clearly stated by the French writer Descartes (about 1635). 
It was also discovered independently by Euler (1752), and 
is often known as Euler^s Theorem. 

The proof of this law is too difficult to be given at this time, but the 
student will be asked to verify it for certain special cases on page 350. 



























§§ 411-413 


REGULAR POLYHEDRONS 


349 


Proposition 16. Regular Polyhedrons 

413. Theorem. There cannot he more than five regular 
polyhedrons. 

Proof. A polyhedral Z has at least three faces. § 361 

Also, the sum of its face A is less than 360°, § 13 

because the polyhedral Z. would flatten out into 
a plane at 36(f. 

Since each Z of an equilateral A is 60°, § 65 

polyhedral A may be formed with three, four, or five equi¬ 
lateral A as faces. . 

Now the sum of six face Z of 60° is 360°. Ax. 1 

Hence not more tha^i ^hY’ee regular polyhedrons with 
equilateral A as face^are possible. 

Since each Z of a square is 90°, § 15 

a polyhedral Z may be formed with three squares as faces. 

Now the sum of four face A of 90° is 360°. Ax. 1 

Hence not more than one regular polyhedron with squares 
as faces is possible. 

Since each Z of a regular pentagon is 108°, § 96 

a polyhedral Z may be formed with three regular penta¬ 
gons as faces. 

Now the sum of four face A of 108° is 432°. Ax. 1 

Hence not more than one regular polyhedron with regu¬ 
lar pentagons as faces is possible. 

Now the sum of three Z of a regular hexagon is 360°; 
of a regular heptagon, more than 360°; and so on. 

Hence there cannot be more than five regular poly¬ 
hedrons. 

It is not of enough importance to prove that there are actually five 
regular polyhedrons as stated in §410. In elementary work the fact 
may be safely assumed. 


350 


GENERAL POLYHEDRONS 


BOOK VII 


Exercises. Polyhedrons 


1. Count the number of edges, vertices, and faces on 
each of the five regular polyhedrons and then fill in the 
blank spaces in the following table: 


Name 

Edges 

Vertices 

Faces 

Tetrahedron . . . 

6 

4 

4 

Hexahedron (cube) . 




Octahedron .... 




Dodecahedron . . . 




Icosahedron .... 





2. From the above table show that in each case the law 
e-\-2 = v -{-f holds true. 

3. Assuming that the law in Ex. 2 is true for all poly¬ 
hedrons, prove that a seven-edged polyhedron is impossible. 

4. If the centers of the six faces of a cube are joined, 
what kind of polyhedron is constructed ? Draw the figure 
and prove that any two edges are equal. 

5. If the centers of the four faces of a regular tetrahe¬ 
dron are joined, what kind of polyhedron is constructed? 
Draw the figure and prove any two edges equal. 

6. As in Ex. 4, consider the figure which results from 
joining the centers of the faces of a regular octahedron. 

7. A quartz crystal is in the form of a 
hexagonal prism with a pyramid on one 
of its bases, as here shown. Show that 
the relation stated in Ex. 2 holds true. 

8. A given polyhedron has six vertices and five faces. 
How many edges are there? 












§§ 414-416 


CYLINDERS 


351 


IV. Cylinders 

414. Cylindric Surface. A surface generated by a moving 
straight line which is always parallel to a fixed straight 
line, and touches a fixed curve not in the plane of the fixed 
line, is called a cylindric surface, or a cylindrical surface. 

The moving line is called the generatrix and the fixed curve is called 
the directrix. In the first figure below, ABC is the directrix of the cylin¬ 
dric surface shown. The generatrix in any position is called an element. 

415. Cylinder. A solid bounded by a cylindric surface and 
two parallel plane surfaces cutting all the elements is called 
a cylinder. 



It is evident that all elements of a cylinder are equal. The terms 
bases, lateral area, and altitude are used as with prisms. 

416. Cylinders classified. If the elements of a cylinder 
are perpendicular to the bases, the cylinder is called a 
right cylinder ; if oblique, it is called an oblique cylinder. A 
cylinder whose bases are circles is called a circular cylinder. 

The second figure above shows a right circular cylinder, and the 
third an oblique cylinder. The straight line through the centers of the 
bases of a circular cylinder is called the axis of the cylinder. 

Since a right circular cylinder can be generated by revolving a rec¬ 
tangle about one side as an axis, it is also called a cylinder of revolution. 

























352 


CYLINDERS 


BOOK VII 


Proposition 17. Bases of a Cylinder 
417. Theorem. The bases of a cylinder are congruent 



Given the cylinder c with bases b and h\ 

Prove that b is congruent to b'. 

Proof. Let X, Y, Z be any three points on the perimeter 


of 6, and let XX’, YY', ZZ' be elements. 

Draw XY, YZ, ZX, X'Y', Y'Z', Z’X’. Post. 1 

Then XY is II to X’Y’, and XX' is II to YY’, §§ 338, 414 
Hence XYY’X’ is a O, § 72 

and XY = X’Y’. §76 

Similarly, YZ = Y’Z’ and ZX = Z’X’, 

and hence lYXYZ is congruent to AX'Y'Z'. § 47 


Place b on b’ so that X lies on X', and Y on Y'; then Z, 
which is any third point on the perimeter of b, has one 
and only one corresponding point on the perimeter of 6'. 

Hence the same is true of every point on the perimeter 
of b ; that is, b can be made to coincide with 6'. 

.*. 6 is congruent to 6'. 


§37 










§417 


EXERCISES 


353 


Exercises. Cylinders 

1. Every section of a cylinder made by a plane passing 
through two elements is a parallelogram. 

2. Every section of a right cylinder made by a plane 
passing through two elements is a rectangle. 

3. Any two sections of a cylinder made by parallel planes 
which cut all the elements are congruent. 

• 4. Any section of a cylinder parallel to the base is con¬ 
gruent to the base. 

5. The straight line joining the centers of the bases of a 
circular cylinder passes through the centers of all sections 
of the cylinder parallel to the bases. 

6. If a rectangle revolves about one of its sides, it forms 
a right circular cylinder. 

7. If two right circular cylinders have equal bases and 
equal altitudes, they are congruent. 

8. The locus of points equidistant from a given line is 
a cylindric surface. 

The given line is called the axis of the surface. 

9. The center of any section of a circular cylinder par¬ 
allel to the base is on the axis. 

10. If parallel planes cut all the elements of a cylindric 
surface, the sections thus formed are congruent. 

11. If a section of an oblique cylinder is made by a 
plane parallel to an element, is the resulting figure a 
parallelogram ? Can it be a rectangle ? Give the proofs. 

12. From the center of the upper base of a right circular 
cylinder 4 in. high lines are drawn to the perimeter of the 
lower base. If the diameters of the bases are 6 in., what is 
the length of each line ? 


354 


CYLINDERS 


BOOK VII 


418. Tangent Plane. A plane which contains an element 
of a cylinder, but does not cut the surface, is called a tan¬ 
gent plane. 

From this definition it is. evident that 

A plane passing through a tangent to the base of a cir¬ 
cular cylinder and the element through the point of contact 
is tangent to the cylinder. 

If a plane is tangent to a circular cylinder, its intersection 
with the plane of the base is tangent to the base. 



419. Inscribed Prism. A prism whose lateral edges are 
elements of a cylinder and whose bases are inscribed in 
the bases of the cylinder is called an inscribed prism. 

The first figure above shows an inscribed prism. The cylinder is 
said to be circumscribed about the prism. 

420. Circumscribed Prism. A prism whose lateral faces are 
tangent to the lateral surface of a cylinder and whose bases 
are circumscribed about the bases of the cylinder is called 
a circumscribed prism. 

The second figure above shows a circumscribed prism. The cylinder 
is said to be inscribed in the prism. 






















§§ 418-422 


CYLINDER AS A LIMIT 


355 


421. Transverse Section. A section of a cylinder made by 
a plane that cuts all the elements is called a transverse 
section of the cylinder. 

If the plane is perpendicular to the elements, the section is called a 
right section. 

422. Cylinder as a Limit. From the principles of limits 
studied in plane geometry, and from the nature of inscribed 
and circumscribed prisms, the properties of the cylinder 
stated below may be assumed without proof. 



If a 'prism 'whose base is a regular polygon is inscribed in 
or circumscribed about a circular cylinder, and if the number 
of sides of the prism is indefinitely increased, 

1. The volume of the cylinder is the limit of the volume of 
the prism. 

2. The lateral area of the cylinder is the limit of the lateral 
area of the prism. 

3. The perimeter of any transverse section of the cylinder 
is the limit of the perimeter of the corresponding section of 
the prism. 

As we increase the number of sides of the inscribed or circum¬ 
scribed prism whose base is a regular polygon, the perimeter of the 
base approaches the circle as its limit (§ 303,1). This brings the lateral 
surface of each prism nearer and nearer the cylindric surface. 










356 


CYLINDERS 


BOOK VII 


Proposition 18. Lateral Area of a Cylinder 

423. Theorem. The lateral area of a circular cylinder 
is the product of an element and the perimeter of a right 
section. 



Given c, a circular cylinder; S, the lateral area; e, an ele¬ 
ment ; and p, the perimeter of a rt. section. 

Prove that S = ep. 

Proof. Let a prism whose base is a regular polygon be 
inscribed in c, and let L be the lateral area and p' the perim¬ 
eter of a rt. section. 

Then L = ep'. §376 

If the number of lateral faces is indefinitely increased, 



L-^S, 

§ 422, 2 


p'->p, 

§422,3 

and 

ep'—^ ep. 

§ 301,1 


.*. S=ep. 

§ 301, 2 


424. Corollary. In a cylinder of revolution of lateral area 
S, total area T, altitude h, and radius r, 

S=2irrhy and T=2 7rr{h-\-r). 













§§ 423-426 


AREA AND VOLUME 


357 


Proposition 19. Volume of a Cylinder 

425. Theorem. The volume of a circular cylinder is 
the product of the base and altitude. 



Given c, a circular cylinder; V, the volume; B, the area of 


the base; and h, the altitude. 

Prove that V=Bh. 

Proof. Let a prism whose base is a regular polygon be 
inscribed in c, and let V' be the volume and B' the area of 
the base. 

Then V=^B’h. §395 

If the number of lateral faces is indefinitely increased, 

§422,1 

B'-^B, §303,2 

and B'h-^Bh. §301,1 

.\V=Bh. §301,2 


426. Corollary. Tn a cylinder of revolution of volume V, 

radius r, and altitude h, ^ 21 , 

’ ’ V^irr h. 

Since R = Trr^ (§ 309), then V='Kr^h. This formula and the first one 
of § 424 are those most used in practical work. 










358 


CYLINDERS 


BOOK VII 


Exercises. Cylinders 

1. The diameter of a well is 5 ft. and the water is 8 ft. 
deep. Reckoning 7] gal. to the cubic foot, how many 
gallons of water are there in the well? 

2. When an irregular solid body is placed under water 
in a right circular cylinder 50 cm. in diameter, the level of 
the water rises 30 cm. Find the volume of the body. 

3. How many cubic yards of earth were removed in 
constructing a tunnel which is 140 yd. long and whose 
cross section is a semicircle 18 ft. in radius ? 

4. Find to the nearest 0.01 in. the radius of a hollow 
cylinder 16 in. high and containing 3 cu. ft. 

5. Given that the height of a cylindric container which 
holds 20 qt. is equal to the diameter, find the altitude and 
the diameter. 

6. Given that the area of the lateral surface of a right 
circular cylinder is S, the volume is V, and the altitude is h, 
find two formulas for the radius r. 

7. Given that the circumference of the base of a right 
circular cylinder is C and the altitude is h, find the volume V. 

8. From the formula T=2 7rr(k-l-r) in §424 find the 
value of r. 

Ex. 8 should be omitted unless quadratics have been studied. 

9. Defining similar cylinders of revolution as cylinders 
formed by the revolution of similar rectangles about corre¬ 
sponding sides, prove that the lateral areas of two such 
cylinders are to each other as the squares of the altitudes 
or as the squares of the radii. 

10. Is Ex. 9 true for total areas ? Prove your answer. 

11. Consider Ex. 9 for volumes instead of lateral areas, 
changing the statement as may be necessary. 


§§ 427-429 


CONES 


359 


V. Cones 

427. Conic Surface. A surface generated by a moving 
straight line which always touches a fixed plane curve 
and passes through a fixed point not in the 
plane of the curve is called a conic surface. 

If a pencil is held by the point and the other end 
is allowed to swing round a circle, the pencil will 
generate a conic surface. 

We may also swing a blackboard pointer about 
any point near the middle in such a way that either 
end shall touch a fixed plane curve, and thus generate 
a conic surface. Such a surface is represented in the 
figure here shown. 

The moving line is celled the generatrix, the fixed 
curve the directrix, and the fixed point the vertex. 

The generatrix in any position is called an element 
of the conic surface. 

Since the generatrix is of indefinite length, the conic surface con¬ 
sists of two portions,— one above and the other below the vertex, as 
shown in this figure. These portions are ,ca,lled the upper nappe and 
lower nappe respectively. 

428. Cone. A solid bounded by a conic surface and a plane 
cutting all the elements is called a cone. 

Since the terms base, lateral area, vertex, 
and altitude are used as with pyramids, their 
further definition is unnecessary. 

429. Circular Cones. A cone whose 
base is a circle is called a circular cone. 

The straight line joining the vertex of a 
circular cone and the center of the base is 
called the axis of the cone. 

Each of these figures shows a circular cone. In the one at the left, 
the axis is perpendicular to the base ; in the other it is oblique. 

Many machine parts, such as roller bearings, bevel gears, taper 
spindles, and the like, are made in the form of circular cones or parts 
of such cones. 




















360 


CONES 


BOOK VII 


430. Further Classification of Cones. If the axis of a circu¬ 
lar cone is perpendicular to the base, the cone is called a 
right circular cone ; if oblique, it is called 
an oblique circular cone. 

Since a right circular cone may be 
generated by the revolution of a right 
triangle about one of the sides of the 
right angle, as shov^n in this figure, it is 
also called a cone of revolution. 

The hypotenuse of the triangle corresponds to an element of the 
surface and is called the slant height of the cone. 

We seldom have occasion to measure any except right circular cones. 

431. Conic Section. A section formed by the intersection 
of a plane and the conic surface of a cone of revolution, as 
in the figures below, is called a conic section. 




Fig. 1 Fig. 2 Fig. 3 Fig. 4 Fig. 5 


In Fig. 1 the conic section is two intersecting straight lines. In Fig. 2 
the conic section is a circle, which is discussed in § 432. In Fig. 3 the 
conic section is an ellipse, the form a circle seems to take when looked 
at obliquely. The orbit of a planet is an ellipse. In Fig. 4 the cutting 
plane is parallel to an element, and the conic section is a parabola, the 
path of a projectile in a vacuum. In Fig. 5 the cutting plane is parallel 
to the axis, and the conic section is a hyperbola. 

The general study of conic sections is not a part of elementary 
geometry, but the names of the sections are so frequently met in gen¬ 
eral reading that they should be known. 

































§§ 430-432 


SECTION PARALLEL TO BASE 


361 


Proposition 20. Section Parallel to Base 

432. Theorem. The section of a circular cone made hy 
a plane parallel to the base is a circle. 



Given a circular cone with the section A'B'C'D' made by the 
plane m II to the base. 

Prove that A'B'C'D' is a Q, 


Proof. Let O be the center of the base, and let O' be the 
point in which the axis VO pierces the plane m. 

Let the planes of FO and any elements VA, VB cut the 
base in the radii OA, OB and the plane m in O'A', O'B', 


Then O'A' is II to OA, and O'B' is II to OB. 

Since O'A', O'R'divide VA, FO, F5 proportionally, 
then • A A OF is similar to AA'O'F, 

and A BOV is similar to AB'O'V, 

OA _VO _ OB 
O'A' FO' O'B'' 


Hence 


§338 

§201 

§213 

§205 


Since OA = OB (§ 134,1), then O'A'= O 'B', § 198, 2 

Since A and B are any points on ABCD, then A' and B' 
are any points on the intersection of m and the cone. 

A'^'C'O'isaO. §134,6 


PS 
















362 


CONES 


BOOK VII 


433. Tangent Plane. A plane which contains an element 
of a cone, but does not cut the surface (as in the first figure 
below), is called a tangent plane. 

From this definition, it is evident that 

A plane passing through a tangent to the base of a circular 
cone and the element through the point of contact is tangent 
to the cone. 

If a plane is tangent to a circtdar cone, its intersection 
with the plane of the base is tangent to the base. 



434. Inscribed Pyramid. A pyramid whose lateral edges 
are elements of a cone and whose base is inscribed in the 
base of the cone is called an inscribed pyramid. 

The second figure above shows an inscribed pyramid. The cone is 
said to be circumscribed about the pyramid. 

435. Circumscribed Pyramid. A pyramid whose lateral 
faces are tangent to the lateral surface of a cone and 
whose base is circumscribed about the base of the cone is 
called a circumscribed pyramid. 

The third figure above shows a circumscribed pyramid. The cone is 
said to be inscribed in the pyramid. 









§§433-437 


CONE AS A LIMIT 


363 


436. Frustum of a Cone. The portion of a cone included 
between the base and a section parallel to the base is called 
a frustum of a cone. 

This figure shows a frustum of a cone of 
revolution. 

The base of the cone and the parallel 
section are together called the bases of the 
frustum. 

The terms altitude and lateral area as 
applied to a frustum of a cone, and slant 
height as applied to a frustum of a right 
circular cone, have the same meaning as they do when applied to a 
frustum of a pyramid. As with frustums of regular pyramids, only 
frustums of cones of revolution have a slant height. 

437. Cones and Frustums as Limits. The following proper¬ 
ties of cones and frustums, similar to those given in § 422, 
may be assumed without proof from a study of the figures 
accompanying the statements: 

1. If a pyramid whose base is a regular polygon is in¬ 
scribed in or circumscribed about a circular 
conCy and if the number of lateral faces of 
the pyramid is indefinitely increased, the 
volume of the cone is the limit of the volume 
of the pyramid, and the lateral area of the 
cone is the limit of the lateral area of the 
pyramid. 

2. The volume of a frustum of a cone is the limit of the 
volumes of the corresponding frustums of the inscribed and 
circumscribed pyramids whose bases are reg¬ 
ular polygons as the number of lateral faces 
is indefinitely increased, and the lateral area 
of the frustum of a cone is the limit of the lat¬ 
eral areas of the corresponding frustums of 
the inscribed and circumscribed pyramids. 















364 


CONES 


BOOK VII 


Proposition 21. Lateral Area of a Cone 

438. Theorem. The lateral area of a cone of revolution 
is half the product of the slant height and the circum¬ 
ference of the base. 



Given a cone of revolution; S, the lateral area; C, the cir¬ 
cumference of the base; and /, the slant height. 

Prove that s=ilC. 

Proof. In a regular circumscribed pyramid whose lateral 
area is L and whose perimeter of base is p, L = ^ Ip. § 401 
If the number of lateral faces is indefinitely increased, 


L-^S, §437,1 

and p^C. §303,1 

Then llp-^llC. §301,1 

.•.S=J^C. §301,2 


439. Corollary. In a cone of revolution of lateral area S, 
total area T, slant height I, and radius r, 

S = ttW, and T— 7rr{l r). 

440. Corollary. The theorem of § Jf03 applies to a circular 
cone. 

The necessary changes in wording are obvious. 













§§ 438-442 


LATERAL AREA 


365 


Proposition 22.^ Lateral Area of a Frustum 

441. Theorem. The lateral area of a frustum of a cone 
of revolution is half the product of the slant height and 
the sum of the circumferences of its bases. 



Given a frustum of a cone of revolution; S, the lateral area; 
C and C'f the circumferences of the bases; and /, the slant height. 

Prove that S = ll{CCf 

Proof. Let L be the lateral area of the corresponding frus¬ 
tum of a regular circumscribed pyramid, and let p, p' be the 
perimeters of the bases corresponding to C, C' respectively. 


Then L = il{p-\-p'). §402 

If the number of lateral faces is indefinitely increased, 
L-^S. §437,2 

From § 303,1, we may assume that 
p-\-p'-^C-\-C’. 

Then ll{p-{-p')-^hliC-\-Cf §301,1 

/. S = jl{C §301,2 


442. Corollary. The lateral area of a frustum of a cone 
of revolution is the product of the slant height and the cir¬ 
cumference of a section equidistant from the bases. 











366 


CONES 


BOOK VII 


Proposition 23. Volume of a Cone 

443. Theorem. The volume of a circular cone is one 
third the product of the base and the altitude. 



Given a circular cone; F, the volume; the area of the 
base; and 7z, the altitude. 

Prove that V=\Bh. 

Proof. Let a pyramid whose base is a regular polygon be 


inscribed in the cone. 

Let V' be the volume and B' the area of the base of the 
inscribed pyramid. 

Then v'=\b% §407 

If the number of lateral faces is indefinitely increased, 
V'^V, §437,1 

and B'-^B. §303,2 

Then ^B'h^^Bh. §301,1 

.\V=\Bh. §301,2 


444. Corollary. In a circular cone of volume V, radius r, 
and altitude h, 

V=\irr\. 














§§ 443-445 


VOLUME 


367 


Proposition 24. Volume of a Frustum 

445. Theorem. The volume of a frustum of a circular 
cone is one third the product of the altitude of the frus¬ 
tum by the sum of the areas of the bases and the mean 
proportional between them. 



Given a frustum of a circular cone; F, the volume; 
the areas of the bases; and /z, the altitude. 

Prove that V=^h{B + B’+ V^'). 

Proof. Let C be the volume of the cone from which the 
frustum is cut; and let C' and h' be the volume and alti¬ 
tude respectively of the small cone remaining after the 
frustum is removed. 


Then V=C-C'=lB{h + h')-\B'h'. 


Now 
Solving, h'= - 


§443 

^B\y/^'=h-\-h'\h. §440, Ax. 6 

hVB' 

B-B' 


Vp-Vp' 

Then F= i j^Pfe + (P - P') j • Ax. 5 

Simplifying, F=M(P + P'+V^). 

Since B = Trr^ and B'=Trr'^, the above formula may be written 
V = ^Trh{T^ rr'). 

This subsidiary proposition may be omitted, if desired. 

























368 


CONES 


BOOK VII 


Exercises. Numerical Computations 

Find the lateral areas of cones of revolution, given the slant 
heights and the circumferences of the bases respectively as 
follows: 

1. 4iin.,5i in. 2. 4.7 in., 6.2 in. 3. 3 ft. 6 in., 5 ft. 

Find the lateral areas of cones of revolution, given the slant 
heights and the radii of the bases respectively as follows: 

4. 2.5 in., 2.1 in. 5. 5.8 in., 5.6 in. 6. 3 ft. 3 in., 7 in. 

Find the total areas of cones of revolution, given the slant 
heights and the radii of the bases respectively as follows: 

7. 4 in., 3| in. 9. 16 in., 7 in. 11. 8 ft., 3^^ ft. 

8. 6 in., 3j in. 10. 18 in., 7 in. 12. 14 ft., 7 ft. 

Find the volumes of circular cones, given the altitudes and 
the areas of the bases respectively as follows: 

13. 4| in., 8f sq. in. 15. 6 in., 5.8 sq. in. 

14. 6^ in., 10 sq. in. 16. 8.1 in., 18.8 sq. in. 

Find the volumes of circular cones, given the altitudes and 
the radii of the bases respectively as follows: 

17. 8 in., 4.2 in. 19. 18 in., 7 in. 21. 4.9 in., 2.1 in. 

18. 3 in., 2.1 in. 20. 21 in., 7 in. 22. 10.5 in., 6.2 in. 

23. How many cubic feet are there in a conical tent 
which is 14 ft. in diameter and 9 ft. high ? 

24. How many cubic feet are there in a conical pile of 
earth which is 18 ft. in diameter and 10 ft. high ? 

25. An isosceles triangle whose altitude is 4 in. and whose 
equal sides are each 5 in. revolves about the base. Find the 
volume of the double cone thus formed. 


§445 


FORMULAS 


369 


Exercises. Formulas 

Deducej if possible, formulas for the following, stating the 
impossible cases, if any: 

1. The lateral area of a cone of revolution in terms of 
the radius of the base and the altitude. 

2. The slant height of a cone of revolution in terms of 
the lateral area and the circumference of the base. 

3. The slant height of a cone of revolution in terms of 
the lateral area and the radius of the base. 

4. The radius of the base of a cone of revolution in terms 
of the lateral area and the slant height. 

5. The slant height of a cone of revolution in terms of 
the total area and the radius of the base. 

6. The circumference of the base of a circular cylinder 
in terms of the lateral area and the slant height. 

7. The volume of a cylinder in terms of the altitude 
and the lateral area. 

8. The altitude of a circular cone in terms of the volume 
and the area of the base. 

9. The area of the base of a circular cone in terms of 
the volume and the altitude. 

10. The altitude of a circular cylinder in terms of the 
volume and the radius of the base. 

11. The radius of the base of a circular cylinder in terms 
of the volume and the altitude. 

12. The volume of a cone of revolution in terms of the 
slant height and the radius of the base. 

13. The slant height and the altitude of a cone of revo¬ 
lution in terms of the volume and the circumference of 
the base. 


370 


CONES 


BOOK VII 


Exercises. Theory of the Cone 

1. Every section of a cone made by a plane passing 
through the vertex is a triangle. 

2. The axis of a circular cone passes through the center 
of every section which is parallel to the base. 

3. Defining similar cones of revolution as cones formed 
by the revolution of similar right triangles about corre¬ 
sponding sides, prove that the lateral areas of two similar 
cones of revolution are to each other as the squares of their 
altitudes. 

4. In Ex. 3, consider the case of the total areas. 

5. Consider Ex. 3 with '' slant heights substituted for 
altitudes. 

6. Consider Ex. 3 with ''radii’’ substituted for "alti¬ 
tudes.” 

7. Consider Ex. 3 with respect to volumes instead of 
lateral areas, changing the statement as may be necessary. 

8. If the lateral surface of a cone of revolution is cut 
along one of the elements and unrolled on a plane, show 
that it becomes a sector of a circle. Show also that there 
can be deduced a formula for the area of a sector of a 
circle that shall be the same as the formula for the lateral 
area of a cone. 

9. Deduce a formula for finding the altitude of a frus¬ 
tum of a circular cone in terms of the volume and the areas 
of the bases. 

If the subsidiary proposition of § 445 was not taken, omit Exs. 9 
and 10. 

10. Deduce a formula for finding the altitude of a frus¬ 
tum of a cone of revolution in terms of the volume and 
the radii of the bases. 


§445 


APPLICATIONS 


371 


Exercises. Industrial Problems 


1. A steamer's funnel 4 ft. 8 in. in diameter is built up 
of four equal plates in girth, with a lap at each joint of 
If in. Find one dimension of each plate. 


2. There is a rule for calculating the strongest beam that 
can be cut from a cylindric log, as follows: Trisect the 
diameter ABy and at the points of division 
P, Q erect the Js PC, QD on opposite sides 
of AB, and meeting the circle in C and D. 

Then ADBC is a section of the strongest 
beam. Given that the diameter AB of a log 
is 18 in., calculate the dimensions AD and 
DB of the strongest beam that can be cut from the log. 

3. A tubular boiler has 128 tubes each 3| in. in diameter 
and 16 ft. long. Find the area of the total surface of the 
tubes to the nearest square foot. 



4. In a room of a factory heated by steam pipes, there 
are 280 ft. of 2-inch pipe, 36 ft. of 3-inch pipe, and 6 ft. of 
4|-inch feed pipe. Find the total heating surface to the 
nearest square foot. 

5. A triangular plate of wrought iron | in. thick is 
3 ft. 4 in. on each side. If the weight of a plate 1 ft. square 
and I in. thick is 5 lb., what is the weight, to the nearest 
pound, of the triangular plate ? 


6. A cylinder 18 in. in diameter is required to hold 60 gal. 
of water. Allowing 231 cu. in. to the gallon, what must be 
the height of the cylinder to the nearest 0.1 in. ? 

7. The water level of an upright cylindric boiler is 1 ft. 4 in. 
below the top of the boiler. If the cross-section area of the 
boiler is 14 sq. ft, what is the volume of the steam space 
above the water ? 





872 


CONES 


BOOK VII 


8. Allowing 490 lb. per cubic foot, find to the nearest 
0.01 lb. the weight of a steel plate 5 ft. by 4 ft. 6 in. by if in. 

9. Through a steel plate 5 ft. long, 3 ft. 8 in. wide, and 
f in. thick, a porthole 12 in. in diameter is cut. Allowing 
0.29 lb. per cubic inch, find the weight of the finished plate. 

10. A cast-iron base for a column is in the form of a frus¬ 
tum of a regular pyramid. The lower base is a square 26 in. 
on a side, the upper base has one fourth the area of the 
lower base, and the altitude of the frustum is 10 in. If the 
total surface is to be painted, what area must be covered ? 


11. A cylinder head for a steam 
engine has the shape shown in 
this figure. Allowing 41 lb. for 
the weight of a steel plate 1ft. 
square and 1 in. thick, find the 
weight of the cylinder head to 
the nearest 0.1 lb. 



12. A hollow steel shaft 14 ft. long has an exterior diam¬ 
eter of 16 in. and an interior diameter of 9 in. Allowing 
0.29 lb. per cubic inch, find the weight to the nearest pound. 






13. A steel beam in the form here shown is 
16 ft. long. Allowing 0.29 lb. per cubic inch, find 
the weight of the beam to the nearest pound. 

14. How many square feet of tin are re¬ 
quired to make a funnel, if the diameters at 
the top and bottom are to be 32 in. and 16 in. 
respectively, the height is to be 24 in., and 4 sq. in. are 
allowed for waste ? 


15. Find the cost, at $1.35 per square foot, of finishing 
the curve surface of a frustum of a right circular cone 
whose slant height is 12 ft. and the radii of whose bases 
are 3 ft. 6 in. and 2 ft. 4 in. respectively. 














§445 


EXERCISES 


373 


Exercises. Equivalent Solids 

1. When a cube of iron 6 in. on an edge is melted it just 
fills a mold in the form of a right prism whose base is a 
rectangle 8 in. long.and 6 in. wide. Find the height of the 
prism and the difference between its total area and the 
total area of the cube. 

2. A pile of bricks in the form of a regular pyramid 
10 ft. high is repiled in the form of a regular prism with 
an equivalent base. Assuming no loss due to piling the 
bricks a different way, find the height of the prism. 

3. The diameter of a cylinder is 12 ft. and its height is 
6 ft. Find the height of an equivalent right prism, the base 
of which is a square 5 ft. on a side. 

4. If one edge of a cube is e, what is the height h of an 
equivalent right circular cylinder whose radius is r ? 

5. The dimensions of a rectangular parallelepiped are 
a, 6, c. Find the height of an equivalent right circular cyl¬ 
inder which has a for the radius of its base. Also find the 
height of an equivalent right circular cone which has a for 
the radius of its base. 

6. A right circular cylinder 5 ft. in diameter is equivalent 
to a right circular cone 5 ft. in diameter. If the height of 
the cone is 6 ft., what is the height of the cylinder ? 

7. The heights of two equivalent right circular cylinders 
are in the ratio 4:9. If the diameter of the first is 5 ft., 
what is the diameter of the second ? 

8. A frustum of a cone of revolution is 5 ft. high and the 
diameters of its bases are 2 ft. and 4 ft. respectively. Find 
the height of an equivalent right circular cylinder whose 
radius is 5 ft. 

Omit Ex. 8 if § 445 was not taken. 


374 


CONES 


BOOK VII 


Exercises. Miscellaneous Problems 

1. The slant height of a frustum of a regular pyramid 
is 24 ft., and the bases are squares whose sides are 50 ft. 
and 20 ft. respectively. Find the volume. 

Exs. 1-6 should be omitted if §§ 408 and 445 were not taken. 

2. Given that the bases of a frustum of a pyramid are 
regular hexagons whose sides are 2 ft. and 3 ft. respec¬ 
tively, and that the volume is 16 cu. ft., find the altitude. 

3. From a right circular cone whose slant height is 24 ft. 
and the circumference of whose base is 8 ft., there is cut off 
by a plane parallel to the base a cone whose slant height is 
5 ft. Find the lateral area and the volume of the frustum. 

4. Find the difference between the volume of a frustum 
of a pyramid, whose altitude is 8 ft. and whose bases are 
squares 16 ft. and 12 ft. respectively on a side, and the vol¬ 
ume of a prism of the same altitude whose base is a section of 
the frustum parallel to its bases and equidistant from them. 

5. A stone windmill is in the shape of a frustum of a 
right circular cone. The mill is 14 m. high, the outer diam¬ 
eters at the bottom and the top are 18 m. and 14 m., and 
the inner diameters are 14 m. and 8 m. respectively. How 
many cubic meters of stone were required to build it? 

6. A brick chimney has the shape of a frustum of a 
regular pyramid. The chimney is 160 ft. high, its upper 
and lower bases are squares 9 ft. and 14 ft. on a side respec¬ 
tively, and a square flue 6 ft. on a side runs from top to 
bottom. How much brickwork does the chimney contain ? 

7. Two right triangles whose bases are 5 in. and 7 in., 
and whose hypotenuses are 8^ in. and Ilf in. respectively, 
revolve about their third sides. Find the ratio of the total 
areas of the solids generated and find their volumes. 


BOOK VIII 


THE SPHERE 

1. Fundamental Theorems 

446. Sphere. The locus of points in space at a given dis¬ 
tance from a given point is called a sphere. 

The terms center, radius, chord, and diameter are used as in the 
case of the circle. 

This is the modern definition, analogous 
to the modern definition of a circle, but it 
will be found that no confusion will arise 
if the student considers a sphere as the 
solid inclosed by a spherical surface or 
spheric surface. In modern mathematics 
the volume of a sphere is considered to 
mean the volume inclosed by the surface 
which is called a sphere. 

447. A Point and a Sphere. A point may be on a sphere, 
within the sphere (inclosed by it), or outside the sphere (not 
inclosed by it). 

448. Properties of a Sphere. As in § 134, we have 

1. All radii of the same sphere or of equal spheres are 
equal. 

2. All spheres of equal radii are equal. 

3. A point is within, on, or outside a sphere according as 
the distance of the point from the center is less than, equal 
to, or greater than the radius. 

There are also other properties, such as that a diameter is twice a 
radius, which are too obvious to require mention. 

375 




376 


FUNDAMENTAL THEOREMS 


BOOK VIII 


Proposition 1. Plane Intersecting a Sphere 

449. Theorem. If a plane intersects a sphere^ the line 
of intersection is a circle. 



Given the plane p intersecting the sphere s with center 0. 
Prove that the line of intersection is a O. 

Proof. Let OA, OB be any two radii of s to points on the 
line of intersection, and let OC be the _L from O to p. 


Draw 

CA and CB. 

Post. 1 

Then 

A AGO and BCO are rt. A, 

§316 


OC = OC, 

Iden. 

and 

OA = OB. 

§ 448,1 

Hence 

AAOC is congruent to ABOC, 

§71 

and 

CA = CB. 

§38 


Then any points A and B, and hence all points on the line 
of intersection, lie on a O. § 134, 6 

That is, the line of intersection is a O. 

450. Great Circle. The line of intersection of a sphere and 
a plane passing through the center is called a great circle 
of the sphere. 














§§ 449-460 CIRCLES OF A SPHERE 377 

451. Corollary. Through any two points on a sphere an 
arc of a great circle can he drawn. 

For the two points and the center of the sphere determine a plane. 

452. Small Circle. The line of intersection of a sphere and 
a plane which does not pass through the center is called 
a small circle of the sphere. 

453. Corollary. Through any three points on a sphere one 
and only one circle can he drawn. 

454. Corollary. A diameter of a sphere perpendicular to 
the plane of a circle of the sphere passes through the center 
of the circle. 

455. Poles of a Circle. If a diameter of a sphere is perpen¬ 
dicular to the plane of a circle of the sphere, the ends of the 
diameter are called the poles of the circle. 

456. Spherical Distance. The length of the smaller arc of 
the great circle joining two points on a sphere is called the 
spherical distance between the points, or, where no con¬ 
fusion is likely to arise, simply the distance. 

457. Polar Distance. The spherical distance from the 
nearer pole of a circle to any point on the circle is called 
the polar distance of the circle. 

The spherical distance of a great circle from either of its poles may 
be taken as the polar distance of the circle. 

458. Quadrant. One fourth of a great circle is called a 
quadrant. 

459. Tangent Lines and Planes. A line or plane that has 
one and only one point in common with a sphere, however 
far produced, is said to be tangent to the sphere, and the 
sphere is said to be tangent to the line or plane. 

460. Tangent Spheres. Two spheres which have one and 
only one point in common are said to be tangent. 


PS 


378 


FUNDAMENTAL THEOREMS 


BOOK VIII 


Proposition 2. Equal Polar Distances 

461. Theorem. All 'points on a circle of a sphere are 
equidistant fro'm either pole of the circle. 



Given A, By C, any points on a O of the sphere 5 ; P, P\ 
the poles of O ARC; and PAP\ PBP', PCP\ arcs of great © 
through Ay By C. 

Prove that arc PA = arc PB = arc PC, 
and that arc P'A — arc P'B= arc P'C. 

Proof. Draw PP' and the chords PA, PB, PC. 

Then PP' is _L to the plane of O ABC, § 455 

and passes through the center of QABC. § 454 

Then chord PA = chord PR= chord PC. § 326, 2 

Since the planes of great © pass through the center 
of s, the ® have equal radii (§ 448,1), and hence are equal. 

. ‘. arc PA = arc PB = arc PC. § 140 

Similarly, by drawing chords from P'to A, B, C, we have 
arc P'A = arc P'B= arc P'C. 

462. Corollary. The polar distance of a great circle is a 
quadrant. 












§§ 461,462 EQUAL POLAR DISTANCES 

Exercises. Circles of a Sphere 

Draw figures to illustrate each of the following: 


379 


1. Great circle. 

2. Small circle. 

3. Poles of a circle. 


4. Spherical distance. 

5. Polar distance. 

6. Quadrant. 


Prove the important properties in Exs. 7-11: 

7. Parallel circles of a sphere have the same poles. 

8. All great circles of a sphere are equal. 

9. Every great circle bisects the sphere. 

10. Two great circles bisect each other. 

11. If the planes of two great circles are perpendicular to 
each other, each circle passes through the poles of the other. 

12. A radius of a sphere perpendicular to a chord of the 
sphere bisects the chord. 

13. Equal chords of a sphere are equidistant from the 
center. 

14. Of the chords of a sphere through a point within it, 
the chords which are perpendicular to the diameter of the 
sphere through the point are the shortest. 

The student has doubtless noticed the analogy between the propo¬ 
sitions relating to the circle and those relating to the sphere. If we 
think of a plane as passing through the center of a sphere and any 
other point under discusssion, the figure of the corresponding propo¬ 
sition in plane geometry will often appear on this plane. 

15. An infinite number of spheres can pass through two 
given points, and their centers lie in a fixed plane. 

16. An infinite number of spheres can pass through thr^e 
given points, and their centers lie on a fixed straight line. 

17. The distance between the centers of two spheres of 
radii r and r' is d. State the condition under which the 
spheres intersect. 


380 


FUNDAMENTAL THEOREMS 


BOOK VIII 


Proposition 3. Pole of a Great Circle 

463. Theorem. A point on a sphere which is at the 
distance of a quadrant from each of two other points 
on the sphere^ not the ends of a diametery is a pole of the 
great circle passing through these points. 



Given the point P on the sphere s\ the quadrants PA, PB; 
and ABCy the great O passing through A and B. 

Prove that Pis a pole of the Q ABC. 

Proof. From O, the common center of the great ©PA, 
PB, APC(§ 450), draw OA, OB, OP. 


Since 

arcs PA and P'B are fourths of great ®, 

§458 

then 

AAOP and BOP are rt. A. 

§171 

Hence 

OP is ± to the plane of Q) ABC, 

§320 

and 

P is a pole of the QABC. 

§455 


« Why may not the two points be the ends of a diameter ? 

464. Spherical Angle. When two great-circle arcs inter¬ 
sect they are said to form a spherical angle. A spherical 
angle is considered as equal to the plane angle formed by 
the tangents to the arcs at their point of intersection. 


















§§ 463-468 


TANGENT PLANE 


381 


Proposition 4. Tangent Plane 

465. Theorem. If a plane is perpendicular to a radius 
at its end on the sphere^ the plane is tangent to the sphere. 



Given the sphere s with the plane nz J_ to the radius OP at P. 

Prove that m is tangent to s. 

Proof. Let A be any point except P in m, and draw OA. 

Then, since OA > OP (§ 326,1), A is outside s. § 448, 3 

Then every point in m, except P, is outside s. 

Hence m is tangent to s. § 459 

466. Corollary. If a plane is tangent to a sphere, it is 
perpendicular to the radius drawn to the point of contact. 

467. Inscribed Sphere. If a sphere is tangent to all the 
faces of a polyhedron, the sphere is said to be inscribed in 
the polyhedron, and the polyhedron is said to be circum¬ 
scribed about the sphere. 

468. Circumscribed Sphere. If all the vertices of a poly¬ 
hedron lie on a sphere, the sphere is said to be circum¬ 
scribed about the polyhedron, and the polyhedron is said to 
be inscribed in the sphere. 













382 


FUNDAMENTAL THEOREMS 


BOOK VIII 


Proposition 5. Tetrahedron and Inscribed Sphere 

469. Theorem. A sphere can he inscribed in any given 
tetrahedron. 



Given the tetrahedron ABCD. 

Prove that a sphere can he inscribed in ABCD. 

Proof. The face A of the dihedral A can be bisected. § 103 

These bisectors and the corresponding edges of the 
dihedral A determine planes. §314,1 

These planes bisect the dihedral A (§ 349), and are the 
loci of points equidistant from the faces. . § 359 

Any two of these bisecting planes, as ABQ and ADQ, 
intersect in a st. line AQ. § 315 

This line AQ cuts another bisecting plane, as CDP, in 
some point 0. 

Hence O, the common intersection of the three bisecting 
planes, is equidistant from the four faces of ABCD. 

Since the A from 0 upon the faces are equal, the faces 
of ABCD are tangent to a sphere with center 0. § § 446,465 

Hence a sphere can be inscribed in ABCD. § 467 









§§ 469-471 TETRAHEDRON AND SPHERE 383 

Proposition 6. Tetrahedron and Circumscribed Sphere 

470. Theorem. A sphere can he circumscribed about 
any given tetrahedron. 



Given the tetrahedron ABCD. 

Prove that a sphere can be circumscribed about ABCD. 

Proof. Let P be the center of the O circumscribed about 
face ABC, and Q the center of the O about face ABD. § 188 
Let PR be J_ to face ABC, and QS to face ABD. § 324 
ThenPJ^ is the locus of points equidistant from A, B, C; 
and QS, of points equidistant from A, B, D. § 328 

Now PR and QS both lie in the plane _L to AB at its mid¬ 
point (§ 329). If QS is II to PR, it is ± to face ABC (§ 331). 
But this is impossible, since QS is _L to face ABD which 
intersects face ABC. Hence PR and QS intersect, as at 0. 

Since O is equidistant from A, B, C, D, 
then A, B, C, D lie on a sphere. § 446 

Hence a sphere can be circumscribed about ABCD. § 468 

471. Corollary. Through four points not in the same plane 
one and only one sphere can pass. 












384 


FUNDAMENTAL THEOREMS 


BOOK VIII 


Exercises. Review 

1. The intersection of two spheres is a circle whose plane 
is perpendicular to the line which joins the centers of the 
spheres and whose center is on that line. 

2. The four perpendiculars erected at the centers of 
the circles circumscribed about the faces of a tetrahedron 
intersect in the same point. 

3. The six planes perpendicular to the edges of a tetra¬ 
hedron at their midpoints intersect in the same point. 

4. The six planes which bisect the six dihedral angles 
of a tetrahedron intersect in the same point. 

5. Circles on the same sphere which have equal polar 
distances are equal. 

6. Equal circles on the same sphere have equal polar 
distances. 

7. Find the locus of points in a plane at a given distance 
from a given point. Also find the locus of such points in a 
three-dimensional space. 

8. A line tangent to a great circle of a sphere lies in the 
plane tangent to the sphere at the point of contact. 

9. Any line in a tangent plane drawn through the point 
of contact is tangent to the sphere at that point. 

10. Through a given point on a given sphere one and 
only one plane can be passed tangent to the sphere. 

11. Find a point in a plane equidistant from two inter¬ 
secting lines in the plane, and at a given distance from a 
given point not in the plane. Discuss the solution for all 
types of cases. 

12. How many points determine a straight line ? a circle ? 
a sphere ? Prove that two spheres coincide if they have 
this last number of points in common. 


§§ 472,473 


SPHERICAL ANGLE 


385 


Proposition 7. Measure of a Spherical Angle 

472. Theorem. A spherical angle is measured hy the 
arc of the great circle which has the vertex of the angle 
as pole and is included between the arms of the angle. 



Given PA, PB, arcs of great © of the sphere s intersecting at 
P; PA', PB', the tangents to these arcs at P; and the arc AB, 
included between PA and PB, of the great O with P as pole. 

Prove that the spherical A APB is measured by arc AB. 


Proof. Let the planes of the great © PA, PB, AB intersect 
in PO, AO, BO respectively. 


Then 

PB'is ± to PO, 

§147 

and 

OB is J_ to PO. 

§171 

Hence PB' 

' is II to OP (§ 57), and similarly PA' 

is II to OA. 


.*. ZA'PB' = ZAOB. 

§343 

Since 

Z A OP is measured by arc AB, 

§171 

then 

ZA'PB' is measured by arc AP. 

Ax. 5 


Z APB is measured by arc.AP. 

§464 


473. Corollary. A spherical angle has the same measure 
as the dihedral angle formed by the planes of the two circles. 











386 


FUNDAMENTAL THEOREMS 


BOOK VIII 


Exercises. Review 

1. If an arc of a great circle passes through the pole of 
another great circle, the two circles are perpendicular to 
each other. 

2. If two great circles are perpendicular to each other, 
each passes through the poles of the other. 

3. If the first of two great circles passes through the 
poles of the second, then the second passes through the 
poles of the first. 

4. If three lines form a triangle, it is possible (§§193,194) 
to construct four circles tangent to all three lines. Consider 
the analogous case for spheres and for the four planes which 
form a tetrahedron. 

5. All straight lines at a given distance from a given 
point are tangent to a certain sphere. 

6. If two planes which intersect in the line AB touch a 
sphere at the points C and D respectively, the line CD is 
_L to AB in the sense mentioned in § 335; that is, a plane 
can be passed through CD _L to AB. 

7. If two unequal spheres intersect, are the tangents 
to the larger sphere from any point in the plane of their 
common circle shorter than the tangents from that point 
to the smaller sphere ? State the proposition in correct form 
and then prove it. 

8. Given that the two points P, P' are 16 cm. apart, find 
the locus of points that are 10 cm. from P and 12 cm. 
from P'. 

9. If the edges of a tetrahedron are tangent to a sphere, 
the sum of any pair of opposite edges is equal to the sum 
of any other pair. 

Compare this with Ex. 6, page 119. 


§§ 474-476 


SPHERICAL POLYGONS 


387 


11. Spherical Polygons 

474. spherical Polygon. A portion of a sphere bounded 
by three or more arcs of great circles is called a spherical 
polygon. 

The difference between the general nature of a spherical polygon 
and that of a plane polygon is that the former lies on a spherical surface 
and has arcs of great circles as its sides, while the latter lies on a plane 
surface and has segments of straight lines as its sides. 

The terms sides, angles, vertices, diagonal, convex, concave, and con¬ 
gruent are used as with plane polygons. 

475. Spherical Triangle. A spherical polygon of three sides 
is called a spherical triangle. 

A spherical triangle may be right, obtuse, or acute, and may also be 
equilateral, isosceles, or scalene. 

The terms spheric polygon and spheric triangle are also used. 

476. Relation of Polygons to Polyhedral Angles. The planes 
of the sides of a spherical polygon form a polyhedral angle 
whose vertex is the center of the sphere, whose face angles 
are measured by the sides of the polygon, 
and whose dihedral angles have the same 
numerical measure as the angles of the 
polygon. 

Thus, the planes of the sides of the polygon ABCD 
here shown form the polyhedral Z 0-ABCD. The 
face A BOA, COB, • • • are measured by the sides BA, 

CB, • • • of the polygon. The dihedral angle whose edge is OA has the 
same measure as the spherical Z BAD, and so on. 

Hence from any property of polyhedral angles, we may 
infer an analogous property of spherical polygons, and 
conversely. 

Since we have considered only convex polyhedral angles in the 
preceding work, we shal^ consider only convex spherical polygons. 

Because of the relation between polyhedral angles and spherical 
polygons, we shall first consider the former. 



388 


SPHERICAL POLYGONS 


BOOK VIII 


Proposition 8. Two Face Angles 

477. Theorem. The sum of any two face angles of a 
trihedral angle is greater than the third face angle. 



Given the trihedral Z V-XYZ with face Z XVZ > face Z XVY 
or face ZFFZ. 

Prove that ZXFF+ Z YVZ > ZXVZ. 

Proof. In the plane XVZ let ZXVW=ZXVY, and 
through any point D of VW draw a line cutting VX in 
A and VF in C. On FFtake VB=VD. 

Then A, B, C determine a plane. § 314,1 

Since AV=AV, VB = VD, and ZAVB = ZAVD, then 
AAVB is congruent to ZAVD (§ 40), and AB = AD{% 38). 


Now 

AB + BC>AD-\-DC. 

Post. 3 


.\BC>DC. 

Ax. 7 

Then since VB=VD, and VC = VC, 



ZBVOZDVC. 

§127 

Then 

ZAVB + ZBVOZAVD + ZDVC. 

Ax. 7 

Hence 

ZAVB + ZBVOZ AVC; 

Ax. 5 

at is. 

ZXVY+Z YVZ >ZXVZ. 









§§ 477,478 


POLYHEDRAL ANGLES 


389 


Proposition 9. Sum of Face Angles 

478. Theorem. The sum of the face angles of a poly¬ 
hedral angle is less than four right angles. 



Given the polyhedral ZF with the face Aa^ by c, , 

Prove that a + 6 + c H— • < 4 rt Z. 

Proof. Let a plane cut all the edges of ZF, thus forming 
the polygon ABC • • •, and let P be any point within ABC • • •. 

Drawing PA, PB, PC, • • •, there are as many ^(PAB, 
PBC, • • •) as there are faces {VAB, VBC, • • •)• 

Hence the sum of the A of all the A with vertex V is 
equal to the sum of the A with vertex P, § 65, Ax. 1 

Now AEAV-\-ABAV>ZBAE, 

ZVBA-\-ACBV>ZCBA,- §477 

Hence the sum of the A at the bases of the A with 
vertex V is greater than the sum of the A at the bases of 
the A with vertex P. Ax. 8 

.*. a + 64-c+ • • • <ZAPB+ZBPC-\-ACPD-\- • • •, Ax.8 
or a + 6 + c H— • < 4 rt. Z. § 13 








390 


SPHERICAL POLYGONS 


BOOK VIII 


Proposition 10. Side of a Triangle 

479. Theorem. Any side of a spherical triangle is less 
than the sum of the other two sides. 



Given the spherical l\ABC, 

Prove that CA < AB -f BC. 

Proof. Let 0-ABC be the corresponding trihedral Z. 

Then Z CO A < Z BOA + Z COB. § 477 

.\CA<AB + BC. §476 

Exercises. Spherical Triangles 

1. Explain how you could proceed to bisect a given 
great-circle arc. 

2. Explain how you could determine the arc that bisects 
a given spherical angle. 

3. Draw a sphere and upon it draw freehand a spherical 
A ARC. With A, R, C as poles draw freehand three great 
circles and show that these circles divide the sphere into 
eight triangles. 

Assume that the center, diameter, and radius are given. ’ 

4. Make a drawing of a sphere and on the sphere show 
an equilateral spherical triangle, each side of which is 90°. 
Then draw a triangle with the three vertices as poles. 








§§ 479-481 


POLAR TRIANGLE 


891 


Proposition 11. Sum of Sides 

480. Theorem. The sum of the sides of a spherical 
polygon is less than 360°, 



Given the spherical polygon ABCD, 

Prove that AB^BCCD-\-DA< 360°. 

Proof. Let 0-ABCD be the corresponding polyhedral Z. 

Then Z5OA-[-ZCO5-bZZ)OC+ZZ)OA<360°. §478 

.-. AB-^BC+CD + DA<m\ § 476 

481. Polar Triangle. The triangle formed by the arcs of 
great circles of which the vertices of a given triangle are 
poles is called the polar triangle of the 
given triangle. 

Thus, if A is the pole of the great circle of 
which a' is an arc, B is the pole of the great circle 
of which b' is an arc, and C is the pole of the great 
circle of which c' is an arc, then AA'B'C' is the 
polar triangle of A ABC. 

If, with A, B, C as poles, entire great circles are drawn, these circles 
divide the sphere into eight spherical triangles. Of these eight tri¬ 
angles, that one is the polar of A ABC whose vertex A\ corresponding 
to A, lies on the same side of BC as the vertex A\ and similarly for 
the other vertices. 

While it is desirable to have a spherical blackboard on which the 
student can draw figures, any small ball will serve the purpose. 








392 


SPHERICAL POLYGONS 


BOOK VIII 


Proposition 12. Reciprocal Polar Triangles 

482. Theorem. If one spherical triangle is the polar 
triangle of another^ then the second is the polar tri¬ 
angle of the first. 



Given ARC, a spherical A, and A'R^C', its polar A. 

Prove that ABC is the polar A of A'B'C'. 

Proof. Since A is the pole of arc B'C\ 

and C is the pole of arc A'B\ § 481 

then R' is at the distance of a quadrant from A and C. § 462 
.'. R' is the pole of arc AC. § 463 

Similarly, A' is the pole of arc RC, 

and C' is the pole of arc AB. 

ABC is the polar A of A'B'C'. § 481 

The student should notice that we may just as well start with ABC 
as the polar triangle of A'B'C', and then prove that AR'C' is the polar 
triangle of ABC. 

It should also be noticed that it is not necessary that either of the 
triangles should be wholly within the other. For example, if we draw 
the figures freehand, taking AB as about 100°, AC as about 100°, and 
BC as about 30°, one triangle will overlap the other. 












§§ 482,483 


POLAR TRIANGLES 


393 


Proposition 13. Angle and Side Supplementary 

483. Theorem. Any angle of either of two 'polar tri¬ 
angles is the supple'ment of the opposite side of the other. 



Given the polar A ABC and 

Prove that A A and a'\ /-B and b'; /LCand c'; ZA' and a; 
^B' and b; Z.C' and c are respectively supplementary. 

Proof. Let arcsA^ and AC produced meet arc^'C' at 


D and E respectively. 

Since B' is the pole of arc AE, 

arc5'£' = 90". §462 

Similarly, arc DC'= 90®. 

Hence arc B'D-\- arc DE + arc DC = 180°, Ax. 1 

or arc DE^a' = 180°. Ax. 5 

But arc DE is the measure of Z A. § 472 

ZA + a' = 180°. Ax. 5 


Similarly, Z5+6'=180°, and ZC+c=180°; henceZAand 
a'; ZB and b'; ZC and c’ are respectively supplementary. 

In a similar way, by considering the angles of A A'R'C' and producing 
the sides of A ABC, the other relations can be proved. 


PS 








394 


SPHERICAL POLYGONS 


BOOK VIII 


Proposition 14. Sum of the Angles of a Triangle 

484. Theorem. The sum of the angles of a spherical 
triangle is greater than 180° and less than 5U0°, 



Given the spherical A ABC. 

Prove that 180° < ZA + Z B + ZC < 540°. 

Proof. Let A'B'C' be the polar A of ABC, with the sides 
of both A lettered as usual. 


Then ZA+a'= 180°, ZB+6'= 180°, ZC+c =180°. §483 



. . Z A -f~ Z B -}- Z C d — 540 , 

Ax.l 


ZA + ZB+ZC = 540°- (a'+ b'-\- d). 

Ax. 2 

Now 

a'+6'+c'<360°. 

ZA + ZB + ZC>180°. 

§480 

Also, 

a'A b'-{- c' ^ 0 . 

ZA + ZB + ZC<540°. 



485. Triangles classified as to Right Angles. Since (§ 484) 
a spherical triangle may have two or even three right 
angles, and two or even three obtuse angles, we find it 
convenient to speak of a spherical triangle which has two 
right angles as birectangular, and one which has three right 
angles as trirectangular. 









§§ 484,485 


ANGLES OF A TRIANGLE 


395 


Exercises. Spherical Polygons 

1. Two sides of a spherical triangle are respectively 
83“ 48' and 64° 59'. What is known concerning the number 
of degrees in the third side ? 

2. Three sides of a spherical quadrilateral are respect¬ 
ively 87° 39', 74° 48', and 68° 56'. What is known (§ 480) con¬ 
cerning the number of degrees in the fourth side ? 

3. If two sides of a spherical triangle are quadrants, the 
third side measures the opposite angle. 

4. In a birectangular spherical triangle the sides oppo¬ 
site the right angles are quadrants, and the side opposite 
the third angle measures that angle. 

Since the angles are right angles, what two planes are perpendicular 
to a third plane ? What two arcs must therefore pass through the pole 
of a third arc ? Then what two arcs are quadrants ? How is the third 
angle measured ? 

5. Each side of a trirectangular spherical triangle is a 
quadrant. 

6. Three planes passed through the 
center of a sphere, each perpendicular to 
the other two, divide the spherical sur¬ 
face into eight congruent trirectangular 
triangles. 

Find the number of degrees in the sides of a spherical 
triangle, given the angles of its polar triangle as follows: 

7. 83°; 78°; 64°. 8. 84°50'; 49°38'; 104°40'. 

Find the number of degrees in the angles of a spherical 
triangle, given the sides of its polar triangle as follows: 

9. 69° 42'38"; 93° 48'8"; 38° 36'15". 

10. 72° 48' 26"; 104° 38' 43"; 90°. 




396 


SPHERICAL POLYGONS 


BOOK VIII 


486. Symmetric Spherical Triangles. If through the center 
0 of a sphere the diameters AA\ BB\ CC' are drawn, and 
the points A, B, C and also the points A', 

B\ C' are joined by arcs of great circles, 
the spherical A ABC and A'B'C' are called 
symmetric spherical triangles. 

In the same way we may form two symmetric 
polygons of any number of sides. We may then 
place the symmetric polygons thus formed in any 
position we choose upon the sphere. 

487. Relation of Symmetric Triangles. Two symmetric tri¬ 
angles are mutually equilateral and mutually equiangular; 
but in general they are not congruent, since they cannot be 
made to coincide by superposition. Thus, in the above figure, 
if the A ABC is made to slide on the sphere until the vertex 
A falls on A\ it is evident that the two triangles cannot be 
made to coincide since, looked at from the point O, the 
corresponding parts of the triangles occur in reverse order. 

The relation of two symmetric spherical triangles, which is similar 
to that of a pair of gloves, may be illustrated by cutting the triangles 
out of the peel of an orange. 



488. Symmetric Isosceles Triangles. Consider, however, 
the case of the symmetric A ABC and A'B'C' in which 
AB = ACy and A'B' = A'C'; that is, 
the two symmetric triangles are 
isosceles. Then because AB, AC, 

A'B', and A'C' are all equal, and 
the A A and A' are equal since they 
were originally formed by vertical 
dihedral angles (§ 486), the two triangles can be made to 
coincide. 



If two symmetric spherical triangles are isosceles, they are 
§uperposable and hence are congruent. 


SYMMETRIC TRIANGLES 


397 


Proposition 15. Symmetric Triangles 

489. Theorem. Two symmetric spherical triangles are 
equivalent. 



Given the symmetric spherical A ARC and A'B'C, 

Prove that A ABC is equivalent to AA’B'C'. 

Proof. Let the A ABC and A'B'C' be placed with their 
corresponding vertices opposite each to each with respect 
to the center of the sphere. Post. 5 

Let P be the pole of the small O through A, B, C\ P\ 
the pole of the small G through A\B\C'; and PA, PP, PC, 
P'A\ P'B\ P'C\ the arcs of great ©. 

Now A PC A and P'C'A' are symmetric. § 486 

Also, arc PA = arc PB = arc PC, 

and arc P'A' = arc P'B' = arc P'C'. § 461 

Then APCA is congruent to AP'C'A' (§ 488), and, similarly, 
APAB, P'A’B' and APBC, P'P'C'are respectively congruent. 

Now A APC = APCA + APAB A APBC, 

and AA'B'C' = AP'C'A’ A-AP'A'B' -\-AP'B'C\ Ax. 10 

.*. A ABC is equivalent to AA'P'C'. Ax. 5 

If the pole Plies outside A ABC, then P'lies outside A A'B'C', and 
each triangle is equivalent to the sum of two symmetric isosceles tri¬ 
angles diminished by a third. Hence the result is the same as before. 









398 


SPHERICAL POLYGONS 


BOOK VIII 


490. Congruent and Symmetric Triangles. In Book I we 
studied the congruence of triangles because upon these 
relations the whole structure of plane geometry is built. 
There are corresponding propositions relating to spherical 
triangles and to trihedral angles, but we do not need them 
for the work in the measurement of the sphere. They will 
therefore be omitted at this point, but will be considered 
in §§ 535-538, where they may be studied if the opportunity 
permits. 

Exercises. Review 

1. If two great-circle arcs intersect, the vertical angles 
are equal. 

2. Every point lying on a great circle which bisects a 
given arc of another great circle at right angles is equi¬ 
distant (§ 456) from the ends of the given arc. 

3. From the center of a sphere three radii, each perpen¬ 
dicular to the other two, are drawn. Find the number of 
degrees in the sides and angles of the spherical triangle 
determined by the ends of the radii. 

4. Is it possible to have a spherical triangle with angles 
of 75°, 80°, and 120°? with angles of 82°, 96°, and 2°? with 
angles of 110°, 80°, and 5° ? with angles of 200°, 150°, and 
190° ? with angles of 188°, 206°, and 250° ? State the reason 
in each case. 

5. The face angles of a polyhedral angle are 90°, 90°, and 
90°. State all that you can with respect to the sides 'and 
angles of the corresponding spherical triangle. 

6. Of what kind of spherical triangle can it be said that 
the triangle is its own polar triangle ? 

7. Draw freehand a spherical triangle with angles of 
200°, 90°, and 90°. 


§§ 490-493 


MENSURATION 


399 


III. Mensuration 

491. Important Measurements. In measuring small tracts 
of land it is customary to consider the earth as flat, since 
the results are sufficiently close for practical purposes. 
When, however, we have to measure large areas, like states 
or countries, it is necessary to make allowance for the fact 
that the earth is a sphere. 

The most important measurements that we need to con¬ 
sider for such purposes are the lengths of lines on a sphere 
and the areas of spherical polygons. We also need to know 
that a great-circle arc is the shortest line on a sphere from 
one point to another, and to know how to find areas. 

For the study of mechanical and astronomical problems 
we need to be able to make a few other measurements, 
including that of the volume of a sphere. 

492. Area of a Sphere. Two formulas relating to the sphere 
are often learned in arithmetic or in algebra. One is the 
formula for the area; namely, 

A =4: tttI 

Since is the area of a great circle on a sphere of 
radius r, this formula states the remarkable fact that the 
area of the sphere is four times the area of a great circle; 
that is, that the area of a certain curve surface is exactly 
that of a certain plane surface. We shall later prove that 
this formula is correct. 

493. Volume of a Sphere. The second formula that may 
already be familiar to the student relates to the volume 
of a sphere; namely, y_4^^3^ 

It is impossible to give a satisfactory explanation of this 
formula in arithmetic or in algebra, but we shall now be 
able to give one that is fairly so. 


400 


MENSURATION 


BOOK VIII 


Proposition 16. Shortest Path between Points 

494. Theorem. The shortest path on a sphere between 
two points is the arc, less than a semicircle^ of the great 
circle joining the two points. 



Given AB^ the arc, less than a semicircle, of the great O 
joining the points A and B on the sphere 5. 

Prove that AB is the shortest path on the sphere between 
A and B. 

Proof. Let C be any point on arc AB, and let XCY and 
RCQ be arcs of the small © which have A and B respec¬ 
tively as poles. 

Then if Y is any point except C on arc XCY, and if A F 
and BY are arcs of great ®, we have 

AY=AC. §461 

Now AYA-BY>AC^-BC. §479 

Taking away A Y from the left member of the inequality 
and AC, its equal, from the right member, we have 
BY>BC. 

BC = BQ. 

.-. BY>BQ. 


Now 


Ax. 7 
§461 
Ax. 5 









§§ 494,495 


GEODETIC LINE 


401 


Hence Y lies outside the O whose pole is B, § 134, 6 

Then, since Y is any point except C on arc XCY, the arcs 
XCY and RCQ have only the point C in common. 

Now let AXRB be any line, which does not pass through 
C, from A to 5 on the sphere. 

This line cuts the arcs XCY and RCQ in the separate 
points X and R ; and if we revolve the line AX about A as 
a fixed point until X coincides with C, we shall have a line 
from AtoC equal to the line AX. 

In like manner, we can have a line from BtoC equal to 
the line BR. 

Hence we can have a line from A to B through C that 
is equal to the sum of the lines AX and BR. 

But such a line is less than the line AXRB by the line XR. 

Hence no line which does not pass through C can be the 
shortest line from AtoB; that is, the shortest path from 
A to 5 is through C. 

But C is any point on the arc AB. 

Hence the shortest line from AtoB passes through every 
point of the arc AB, and consequently coincides with the 
arc AB. 

That is, the shortest path from A to 5 is the arc AB of the 
great O through A and B. 

This fact is of great importance in navigation. 

495. Geodetic Line. Just as we use straight lines in meas¬ 
uring distances on a plane, we use great-circle arcs in 
measuring distances on a sphere. Since these arcs are 
used in geodesy, the science of measuring the earth's sur¬ 
face, they are called geodetic lines. 

If we examine the map on a globe, we see that the geodetic line 
from New York to Plymouth, England, goes much farther north than 
we should think if we looked only at a flat map. 


402 


MENSURATION 


BOOK VIII 


Proposition 17. Surface Generated by a Line 

496. Theorem. If a straight-line segment revolves about 
an axis in its plane, the area of the surface generated is 
the product of the projection of the line segment on the 
axis and the circumference of the circle whose radius is 
the perpendicular to the line segment at its midpoint 
included between this point and the axis. 



Given AB, a segment revolving about an axis XY in the 
same plane; CD, the projection of ARon-YF; M, the midpoint 
of AB ; MR, the -L to AB at M\ and S, the area generated by AB, 

Prove that S=CD • 2 irMR. 

Proof. Let ATO be _L to XY. 

1. If AB is II to XY, CD — AB, MR coincides with MO, and a rt. cylin¬ 
der is generated. Hence S=AB • 2 7rMO= CD • 2TrMR (§ 424). 

2. If AB is not II to XY and does not cut XY, let AE be II to XY. 
In the similar AMOR, AEB (§ 210), MO : AE=MR: AB (§ 205); whence 
AB • MO = AE • MR = CD • MR (§ 198,1). Then, since AB generates a 
frustum of a cone of revolution, S=AB • 27rMO= CD • 2 7rMR <§ 442). 

3. If A lies on XY, CD = AD. In the similar AADB, MOR (§ 210), 
AB:MR = AD:MO (§205); whence AB ’ MO = AD - MR = CD ' MR 
(§ 198,1). Now MO = \BD (§ 87). Then, since AB generates a cone of 
revolution, S = ^ AB • 2 irBD = AB • 2 ttMO = CD • 2 ttMR (§ 438). 

The student should give the proof of each case in complete form. 


















§§496,497 


AREA OF A SPHERE 


403 


Proposition 18. Area of a Sphere 
497. Theorem. The area of a sphere of radius r is 4= irr’^. 



Given a sphere of radius r and area S. 

Prove that 8=4 7^r^ 

Proof. Let the sphere be generated by the revolution of 
the semicircle APB about AB, let AMPNB be a regular 
inscribed semipolygon oi 2 n sides, and let Js from the 
center O be constructed to the equal chords AM, MP, • • *. 
Then these Js bisect the chords (§ 141) and are equal (§ 150). 
Let r' be the length of each of these equal Js, and let MM' 
and NN' be J. to AB, 

Since arc AP= arc BP{^ 140), PO also is J_ to AB, 

Then the area generated by AM is AM' \2 ttt', 

the area generated by MP is M'O • 2 irr', * • *. § 496 
Letting 8' represent the area of the surface generated 
by AMPNB, we have 8' =AB-2 ttt'. Ax. 1 

Now if the number of sides is indefinitely increased, 

rWr, §303,3 

and hence AB • 2 irr'—^ AB • 2 ttt. § 301,1 

But, always, 8' = AB' 2'Trr' = 4irrr', § 496 

and obviously 8'^8. 

8=47rr\ 


§ 301,2 











404 


MENSURATION 


BOOK VIII 


498. Spherical Degree. Just as we take a unit of length, 
commonly 1°, in measuring an arc, so we take a unit of 
area in measuring a spherical figure. This 
unit, called a spherical degree, is a spherical 
triangle of which two sides are quadrants 
and the third side is an arc of 1°. Since 
the triangle is evidently of a hemi¬ 
sphere, we see that 

A spherical degree is of a sphere. 

The angles of the unit triangle are evidently angles of 90°, 90°, and 1° 

499. Lune. A portion of a sphere bounded by the halves 

of two great circles is called a lune. p 

A lune is therefore part of a spherical surface. It 
may, by extending the definition of polygon, be con¬ 
sidered as a polygon of two sides. 

Since the two angles formed by the sides of a lune 
are equal, either angle may be taken as the angle 
of the lune. In this figure, ZAPB may be taken as 
the angle of the lune PAP'B. 

500. Area of a Lune. Since the area of a lune whose 
angle is 1° is twice a spherical degree, and the area of a 
lune increases at the same rate as the angle, we see that 

The number of spherical degrees in the area of a lune is 
twice the number of degrees in the angle of the lune. 

501. Spherical Excess. The number of degrees by which 
the sum of the angles of a spherical polygon of n sides ex¬ 
ceeds (n—2)180 is the spherical excess of the polygon. 

In a spherical triangle, we have n — 2 = 3 — 2 = 1. Hence the spherical 
excess of a triangle is the number of degrees by which the sum of its 
angles exceeds 180. 

For example, if the angles of a spherical triangle are 80°, 90°, and 
100°, the spherical excess of the triangle is 90. If the angles of a 
spherical polygon are 150°, 155°, 90°, 55°, 100°, the spherical excess is 
150 -h 155 4- 90 4- 55 -h 100 - (5 - 2) 180, or 550 - 540, or 10. 






§§ 498-501 


SPHERICAL FIGURES 


405 


Exercises. Areas 

1. The area of a sphere is the product of the diameter 
and the circumference of a great circle. 

2. The areas of two spheres are to each other as the squares 
of their radii, or as the squares of their diameters. 

3. The area of a sphere is 72 sq. in. Find the area of a 
lune whose angle is 10°. 

It is evidently of the area of the sphere. 

4. The area of a sphere is 1440 sq. in. Find the number 
of square inches in a spherical degree. 

5. If the number of square inches in a spherical de¬ 
gree is 15, what is the area of the sphere ? 

6. The area of a lune is 10 sq. cm. and the angle of the 
lune is 36°. Find the area of the sphere. 

7. The area of a lune is two spherical degrees. Find 
the angle of the lune. 

8. The angle of a lune is 15° 30'. Find the area of the 
lune in spherical degrees. 

9. A lune with an angle of 20° is on a sphere which has 
a radius of 7 in. Find the number of square inches in the 
area of the lune. 

In this and the following exercises, take 7r= 3}. 

Find the areas of spheres with radii as follows: 

10. 4.9 in. 11. 3.5 in. 12. 2 ft. 4 in. 13. 6 ft. 5 in. 

Find the areas of spheres with diameters as follows: 

14. 42 in. 15. 5.6 in. 16. 6.3 in. 17. 4 ft. 8 in. 

Find the radii of spheres with areas as follows: 

18. 616 sq. in. 19. 2464 sq. in. 20. 15,400 sq. mm. 


406 


MENSURATION 


BOOK VIII 


Proposition 19. Area of a Triangle 

502. Theorem. The area of a triangle of spherical 
excess E on a sphere of radius r is 



Given ARC, a A on a sphere of radius r; R, the spherical 
excess of the A; and S, the area of the A. 

Prove that S = lio EttP, 

Proof. Let the sides of the A ABC be produced to form 
great ©, and let AA’, BB\ CC’ be diameters. 

Then AAB'C' and A'BC are symmetric. § 486 

AAB'C' is equivalent to A A'BC. § 489 

Hence we have 

R + AAB'C' = lune of ZA = 2A spherical degrees, 

S-h AAR'C= lune of ZR = 2R spherical degrees, 


S-\-AAC'B = lune of Z C = 2C spherical degrees. § 500 
.*. 2 5+J sphere = 2 (A+R+C) spherical degrees. Ax. 1 
or 28 + 360 spherical degrees = 2(A+R+C); Ax.5 

whence 8=A+R+C—180. Ax. 4 

Hence the area of A ABC in spherical degrees is E. § 501 
But each spherical degree is o of 4 irp. § 498 
Hence S = E • 4 irP = y EttP. 


















§§ 502-506 


AREA OF A POLYGON 


407 


503. Corollary. A spherical triangle is equivalent to a lune 
whose angle is half the spherical excess of the triangle. 

In the proof of §502 we showed that C—180, and the 

number of spherical degrees in the area of a lune is twice the number 
of degrees in the angle of the lune (§ 500). 

504. Corollary. In a polygon of area S and spherical excess 
E on a sphere of radius r, 

S = -i\^Ei7r\ 

In the spherical polygon here shown, by drawing all the diagonals 
from any vertex, we have a A with each of the sides as base except 
the two which meet at the vertex; that is, there are (n —2)A. 

Since (§ 502) the area of each A is times the 

excess of the number of degrees in each A over 180, the 
sum of the areas of the A is times the excess 

of the sum of all the A over {n — 2)180. 

But (§ 501) the spherical excess of the polygon is the 
sum of the A less {n — 2) 180, and hence 

505. Corollary. A spherical polygon is equivalent to a 
lune whose angle is half the spherical excess of the polygon. 

This follows from §§ 503 and 504. 

506. Significance of the Formula of § 504. It should be 
noticed that E in the formula of § 504 depends upon the 
number of sides in the polygon, since it is always the num¬ 
ber of degrees in the sum of the angles minus (n —2)180. 

In the case of a lune there are two sides, and hence 
{n — 2) 180 = 0. Thus the spherical excess is simply the sum 
of the two equal angles, or twice the angle of the lune. 

Hence in a lune of ZL, we have if^Z/7rr^= 

If L = 360°, that is, if the lune covers the entire sphere, 
= 4 7rr^ as found in § 497. 

The advantage in using §§ 502 and 504 instead of §§ 503 and 505 is 
apparent, since in the former we have a single formula for the area 
without reference to the lune. 



408 


MENSURATION 


BOOK VIII 


Exercises. Areas of Spherical Polygons 

1. Find the area of a triangle with angles of 110°, 100°, 
and 95° on a sphere with a radius of 6 in. 

We have £■=110 +100 + 95 -180 = 125, 
and S=jIo i. 6 • 6 =78f. 

Hence, the area of the triangle is 78.57 sq. in. 

In all cases take 3y, or for ir except when otherwise directed. 

2. Find the area of a polygon with angles of 100°, 110°, 
120°, and 170°, on a sphere with a radius of 63 in. 

In this case £'=100+110 + 120+170 - 2 X180 =140, 
and S= ’ ¥- • 63 • 63 = 9702. 

Hence the area of the polygon is 9702 sq. in. 

3. Taking the radius of the earth as 4000 mi., find the 
area of the earth. 

4. A triangle on the earth ^s surface has one vertex at 
the north pole and the others on the equator, one at 30° W. 
and the other at 20° E. Considering the earth as a sphere 
with a radius of 4000 mi., find the area of the triangle. 

5. In making a survey of part of a continent a triangle 
was laid out with angles of 60°, 100°, and 20° 6'. Find the 
area of the triangle to the nearest 1000 sq. mi. 

Find the areas of triangles with angles as follows on 
spheres of the given radii: 

6. 130°, 100°, 95°; r = 7 in. 7. 110°, 100°, 40°; r = 35 in. 

Taking ir = 3.1J^16, find the areas of spherical polygons 
with angles as follows on spheres of the given radii: 

8. 136°, 154°, 70°, 90°; r = 16 in. 

9. 145°, 150°, 90°, 100°, 130°; r = 30in. 

10. 175°, 168°, 88°, 142°, 100°, 90°; r = 40in. 


§§ 507-509 


ZONES 


409 


507. Zone. A portion of a sphere included between two 
parallel planes is called a zone. 

If a great circle revolves about its diameter as an axis, 
any arc of the circle generates a zone. 

It must be remembered that sphere means the same as spherical 
surface, so that a zone, like ^ lune, is a surface. Thus on the earth we 
have the torrid zone, which is that part of the earth’s 
surface included between the planes of the tropics 
of Cancer and Capricorn. 

The circles made by the planes are called the 
bases of the zone, and the distance between the 
planes is called the altitude of the zone. 

If one of the planes is tangent to the sphere 
and the other plane cuts the sphere, the zone is 
called a zone of one base. If both planes are tangent to the sphere, 
the zone is a complete sphere. 


P 



508. Corollary. In a zone of area S and altitude h on a 


sphere of radius r, 


S = 2 irrh. 


If we apply the reasoning of § 497 to the zone generated by the 
revolution of arc PN, we have p 

S = ON'x27rr. 

But ON' is the altitude h. 

Hence S=27rrh. 



For example, if the radius is 14 in. and the altitude of the zone is 5 in., 
S=2 TTvh = 2 • •14-5 sq. in. = 440 sq. in. 


509. Areas on a Sphere. The most important formulas for 
the areas on a sphere may be summarized as follows: 


Sphere, 
Triangle, 
Polygon, 
Lune of ZL, 
Zone, 


S=4 7rZ. 

= li'o Eirr^. 
S=^hE'jrr\ 
S = qV Lirr^. 
S=2 irrh. 


PS 








410 


MENSURATION 


BOOK VIII 


Exercises. Areas 

1. The area of a zone of one base is equivalent to the 
area of the circle whose radius is the chord of the gener¬ 
ating arc. 

2. Two zones on the same sphere are to each other as 
their altitudes. 

3. If the earth's radius is 4000 mi., and the altitude of 
the torrid zone is 3200 mi., what is the area of this zone ? 

On a sphere whose radius is IJf in.,, find the areas of lunes 
whose angles are as follows: 

4. 60°. 5. 90°. 6. 42° 30'. 7. 32° 20'. 

8. Find the area of a lune whose angle is 40° on a sphere 
with a radius of 28 in. 

9. Find the area of a lune whose angle is 85° on a sphere 
with a diameter of 21 in. 

Taking 'Tr = Y^ fi'^d the areas of spherical polygons with 
angles as follows on spheres of the given diameters: 

10. 140°, 90°, 60°, 120°; d = 20 in. 

11. 170°, 160°, 95°, 30°, 100°; d = 32in. 

Taking 'rr = 3.1J^16, find the areas of spherical polygons 
with angles as follows on spheres of the given circumferences: 

12. 140°, 120°, 100°, 130°, 100°; C= 6.2832 in. 

13. 120°, 140°, 130°, 80°, 160°, 135°; C = 18.8496 in. 

Taking tt = S.H, find the areas of spherical polygons with 
angles as follows on spheres of the given areas: 

14. 70°, 160°, 90°, 120°; S = 600 sq. in. 

15. 65° 30', 140° 50', 95° 34', 138° 50'; S = 600 sq. in. 


§§ 510-512 


INSCRIBED SPHERE 


411 


510. Sphere Inscribed in a Cube. It is readily seen that 
the lines joining the centers of the opposite faces of a cube 
are each equal to an edge of the cube, 
that they all intersect in one point, 
and that this point is equidistant 
from all six faces. It therefore seems 
obvious that a sphere can be in¬ 
scribed in a cube. 

The formal proof of this fact can easily 
be given, but the student will probably 
see that its truth is so evident as to render a proof unnecessary. 

Similarly, we speak of a sphere as inscribed in a cylinder, but no 
formal definition is necessary. 

511. Volume of a Sphere. The volume inclosed by a sphere 
is called the volume of the sphere. 

512. Sphere as a Limit. Suppose that a plane is tangent 
to a sphere at the point on the sphere determined by the line 
joining any vertex of the circumscribed cube to the center 
of the sphere. It is then apparent that since some of the 
cube has been cut off, the volume of the circumscribed 
polyhedron thus formed is nearer than that of the cube to 
the volume of the sphere. 

This process may be continued for all the vertices of 
the cube, repeated for all the vertices of the circumscribed 
polyhedron thus formed, again repeated, and so on indefi¬ 
nitely. The volume of the circumscribed polyhedron thus 
approaches the volume of the sphere. That is, if V is the 
volume of the sphere, and V' is the volume of the circum¬ 
scribed polyhedron of n faces, as described above, then 

V' V as ^ 00 . 

Similarly, if A is the area of the polyhedron of n faces, 
as described above, and S is the area of the sphere, then 
A-^S as n—yoo. 










412 


MENSURATION 


BOOK VIII 


Proposition 20. Volume of a Sphere 

513. Theorem. The volume of a sphere of radius r is 
I Trrl 



Given a sphere of radius r and volume V, 

Prove that V = 17rrl 

Proof. Let the sphere be inscribed in a cube (§ 510) whose 
edge is 2 r. Then the lines joining the center to the vertices 
of the cube are the edges of six pyramids of altitude r 
whose bases are the faces of the cube. 

One such pyramid is shown in the figure. 

The volume of each pyramid is a face of the cube multi¬ 
plied by ir (§407), and the volume of the six pyramids, or 
of the whole cube, is the area of the surface of the cube 
multiplied by Jr (Ax. 1). 

Now let planes be tangent to the sphere at the points 
where the edges of the pyramids cut the sphere. 

One such plane is shown in the figure. 

We then have a circumscribed solid whose volume V', 
although greater than the volume V of the sphere, is nearer 
V than is the volume of the circumscribed cube. 













§513 


VOLUME OF A SPHERE 


413 


Also the area A of the circumscribed solid is nearer the 
area S of the sphere than is the area of the cube. 

Proceeding as before, let the center of the sphere be 
connected with the vertices of the new polyhedron. These 
connecting lines are the edges of pyramids of altitude r 
whose bases are the faces of the polyhedron. 

Then the sum of the volumes of these pyramids is the 
area of the circumscribed polyhedron multiplied by ^r; 
that is, as before, 

V'=A-ir. 


If we continue to increase indefinitely the number of 
faces of the circumscribed polyhedron, we see that 



y'^Uand A-^S. 

§512 


.\V=S^lr. 

§ 301,2 

But 

S —4: irr^. 

§497 

Hence 

y= 47rr^ • \r, 

Ax. 5 

or 

y= I 7rf\ 



Exercises. Volume of a Sphere 

1. The volume of a sphere is the product of the area of 
its surface and one third of its radius. 

2. The volume of a sphere of diameter d is \ ttcZI 

3. The volumes of two spheres are to each other as the cubes 
of their radii or as the cubes of their diameters. 

4. If the radius of the earth is 4000 mi., and if the atmos¬ 
phere extends 50 mi. above the surface of the earth, what 
is the volume of the atmosphere ? 

5. If a solid iron ball 4 in. in diameter weighs 9 lb., what 
is the weight of a spherical iron shell which is 2 in. thick 
and has an external diameter of 20 in. ? 


414 


MENSURATION 


BOOK VIII 


Exercises. Area and Volume of a Sphere 

1. How many square feet of lead are needed to cover a 
hemispherical dome which is 66 ft. in circumference ? 

The student should always use as the value of tt, unless other¬ 
wise directed. 

2. A hollow ball 8 ft. in diameter surmounts the dome 
of a church. Making no allowance for the support, how 
much will it cost to gild the surface of the ball at 10<f per 
square inch ? 

3. Taking the circumference of the earth as 25,000 mi., 
find the area of the surface to the nearest million square 
miles. 

Find the volumes of spheres whose radii are : 

4. 4.2 in. 6. Tin. 8. 3|-in. 10. 5.6 ft. 

5. 6.3 in. 7. 14 in. 9. 10|-in. 11. 4 ft. 1 in. 

12. The diameter of a spherical basket ball is 10 in. 

Allowing 50 sq. in. for waste, how many square inches of 
leather are needed to cover it ? 

13. The distance across the top of a bowl in the shape 
of a portion of a sphere is 14 in. and the greatest depth 
is 7 in. Allowing gal. to the cubic foot, how many pints 
of water does the bowl hold? 

14. If the numbers expressing the area and the vol¬ 
ume of a sphere are the same and the units of measure are 
the square inch and the cubic inch respectively, what is the 
diameter of the sphere ? 

15. The weights of two spheres are in the ratio 2: 5 and 
the weights of 1 cu. in. of each of the substances of which 
they are composed are in the ratio 7 : 2. Find the ratio of 
the diameters. 


§513 


GENERAL REVIEW 


415 


IV. General Review 
Exercises. Polyhedrons 

1. The lines drawn from each vertex of a tetrahedron 
to the point of intersection of the medians of the opposite 
face meet in a point P which divides each line so that the 
ratio of the shorter segment to the whole line is 1:4. 

The point P is called the center of gravity of the tetrahedron. 

2. In Ex. 1, the lines which join the midpoints of the 
opposite edges of the tetrahedron are each bisected by the 
center of gravity. 

3. The plane which bisects a dihedral angle of a tetra¬ 
hedron divides the opposite edge into segments proportional 
to the areas of the faces including the dihedral angle. 

4. Show how to cut a cube by a plane so that the section 
shall be a regular hexagon. 

5. If the face angles at the vertex of a triangular pyra¬ 

mid are all right angles, if the areas of the lateral faces 
are A, B, and C respectively, and if the area of the base 
is P>, then D\ 

6. Show how to cut a tetrahedron by a plane so that the 
section shall be a parallelogram. 

7. The altitude of a regular tetrahedron is equal to the 
sum of the four perpendiculars drawn from any point 
within the tetrahedron to the four faces. 

8. Draw figures to show how to cut a cube so as to have 
a section of three sides; of four sides; of five sides; of as 
many more sides as possible. 

9. The . section of a regular octahedron made by a plane 
parallel to and midway between any pair of opposite faces 
is a regular hexagon. 


416 


GENERAL REVIEW 


BOOK VIII 


Exercises. Formulas Relating to the Sphere 

Deduce formulas for the following: 

1. The area of a zone of height \r on a sphere of 
radius r. 

In the exercises upon this page, and in similar cases, the unit of 
area is the square of the unit of length, and the unit of volume is the 
cube of the unit of length. That is, if we think of r as feet, then S will 
be square feet and V will be cubic feet. 

2. The volume F of a sphere in terms of C, the circum¬ 
ference ; in terms of S, the area of the sphere. 

3. The radius r of a sphere in terms of V, the volume. 

4. The area S of the zone, on a sphere of radius r, which 
is illuminated by an electric light at the height h above 
the surface. 

5. The diameter d of a sphere in terms of S, the area of 
the sphere. 

6. The altitude of a zone of area 5 on a sphere of 
volume V. 

7. The volume of the metal in a spherical iron shell of 
which the internal radius is r, and the thickness of the 
metal is t. 

8. The weight of a spherical metal shell in which the 
inside radius is r, the thickness of the metal is t, and the 
weight of a cubic unit of the metal is w. 

9. The diameter of the sphere upon which a zone of 
area S has an altitude h. 

10. The area 5' of a zone of altitude h upon a sphere 
of area S. 

11. Hr and r are the radii of the spheres circumscribed 
about and inscribed in a regular tetrahedron of edge 2 a, 
then = 3 r = 2 aV6. 


§513 


SPHERE AND CYLINDER 


417 


Exercises. Cylinders 

1. The volume of a right circular cylinder is the product 
of the lateral area and half the radius. 

2. The volume of a right circular cylinder is the product 
of the area of the rectangle, which generates the cylinder 
by revolving about one of its sides, multiplied by the cir¬ 
cumference of the circle generated by the point of inter¬ 
section of the diagonals of the rectangle. 

3. If the altitude of a right circular cylinder is equal to 
the diameter of the base, the volume is the product of the 
total area and one third the radius. 

4. By what number must the dimensions of a cylinder 
of revolution be multiplied to obtain a similar cylinder 
of revolution (Ex. 9, page 358) whose entire surface area is 
twice the first ? n times the first ? 

5. What is the multiplier in Ex. 4 if the volume of the 
second cylinder is to be twice that of the first ? is to be n 
times that of the first ? 

6. Compare the volumes of the solids generated by the 
successive revolution of a rectangle of base h and altitude 
h about two adjacent sides. 

7. Find the radius of a right circular cylinder in which 
the number of cubic units of volume is equal to the number 
of square units of the area of the entire surface. 

8. Find to the nearest square centimeter the area of the 
total surface of a cylinder whose altitude is 7.6 cm. and the 
diameter of whose base is 4.2 cm. 

9. Find to the nearest cubic centimeter the volume of 
a cylinder whose altitude is 6 cm. and which fits- exactly 
into a right prism whose base is a square that is 1.8 cm. 
on a side. 


418 


GENERAL REVIEW 


BOOK VIII 


Exercises. Cones and Pyramids 

1. The altitude of a cone of revolution is 24 in. and the 
radius of the base is 10 in. Find the radius of the sector of 
paper which, when rolled up, will just cover the convex 
surface of the cone. Find also the number of degrees in 
the angle of this sector. 

The second result may be expressed either in degrees with a decimal 
fraction, or in degrees, minutes, and seconds. 

2. The volume of a regular pyramid is the product of 
one third its lateral area and the perpendicular distance 
from the center of the base to any lateral face. 

3. Find the volume of a pyramid whose base is 30 sq. in. 
and one of whose lateral edges, which makes an angle of 
45° with the plane of the base, is 5 in. long. 

4. A pyramid is cut by a plane parallel to the base and 
bisecting the altitude. What is the ratio of the volume of 
the pyramid cut off to that of the entire pyramid ? 

5. Consider Ex. 4 for the case of a cone. 

6. The height of a regular hexagonal pyramid is 6 in. 
and one edge of the base is 1 in. Find the volume and also 
find the volume of the pyramid cut off by a plane 4 in. 
from the base and parallel to it. 

7. One of the lateral edges of a regular hexagonal pyr¬ 
amid is 6 in., and the radius of the circle circumscribed 
about the base is 1 in. Find the altitude, the volume, the 
lateral area, and the area of the total surface. 

8. If a right triangle of hypotenuse h and sides a and h 
revolves about h as an axis, what is the volume of the 
solid thus generated ? 

9. If the radius of the base of a right circular cone is r 
and the angle at the vertex is 120°, what is the volume ? 


§513 SPHERES, CYLINDERS, AND CONES 419 

Exercises. Spheres, Cylinders, and Cones 

1. The area of a sphere is two thirds the area of the total 
surface of the circumscribed cylinder, 

2. The volume of a sphere is two thirds the volume of the 
circumscribed cylinder. 

Exs. 1 and 2 were discovered by Archimedes, one of the greatest 
mathematicians of Greece, about 250 b.c. 

3. A sphere of radius 6 in. and a right circular cone of the 
same radius stand on a plane. If the. height of the cone is 
equal to the diameter of the sphere, find the position of 
the plane that cuts the two solids in equal circular sections. 

4. In a cylindric jar 8 in. in diameter, water is standing 
to a depth of 6 in. If an iron ball 4 in. in diameter is dropped 
into the jar, what is then the depth of the water ? 

5. On the base of a right circular cone a hemisphere is 
constructed outside the cone. Given that the area of the 
hemisphere is equal to that of the cone, and that the radius 
is r, find the slant height of the cone, the inclination of the 
slant height to the base, and the volume of the entire solid. 

6. Find the area of a sphere inscribed in a cylinder of 
volume i TTcZ®, where d is the diameter of the sphere. 

7. Find the volume of a sphere inscribed in a cylinder 
of area ird{2 -\-d). 

8. A sphere of radius r is inscribed in a cylinder. Find 
the volume of the cylinder not occupied by the sphere. 

9. A cylinder is circumscribed about a hemisphere, and 
a cone is inscribed in the cylinder so as to have its vertex 
on the upper base and to have its base in common with 
the lower base of the cylinder. Prove that the volumes of 
the cone, the hemisphere, and the cylinder are proportional 
to 1, 2, 3. 


420 


GENERAL REVIEW 


BOOK VIII 


Exercises. Portions of the Surface of a Sphere 

1. If the altitude of the north temperate zone is 1800 mi., 
what is the area of the zone ? 

In Exs. 1-7 take 4000 mi. as the radius of the earth. 

2. How far in one direction can a man see from an 
ocean steamer if his eye is 50 ft. above the water ? 

3. How many square miles of the earth's surface can 
be seen from an airplane at an elevation of 10,000 ft. ? 

4. At what height above the earth must a man be in 
order to see one eighth of the surface ? 

5. What fractional part of the earth’s surface could be 
seen if an observer were at the height of the earth’s radius 
above the sea ? 

6. If the ocean area is three fourths of the earth’s sur¬ 
face and the average depth of the water is 2 mi., what is 
the volume of water in the oceans ? 

7. In a lighthouse on an isolated rock the light is placed 
168 ft. above the surface of the sea and can be seen from 
any point within a circle reaching to the horizon. Find the 
number of square miles of the earth’s surface inclosed by 
this circle. 

Find the areas of triangles with angles as follows on 
spheres of the given radii: 

8. 120°, 110°, 90°; r = 7 in. 10. 120°, 95°, 90°; r = 14 in. 

9. 105°, 105°, 80°; r = 7 in. 11. 90°, 90°, 90°; r = 91 in. 

Find the areas of polygons with angles as follows on 
spheres of the given radii: 

12. 140°, 150°, 80°, 80°; r = 14 in. 

13. 120°, 100°, 185°, 80°, 100°; r = 21 in. 


§513 


THE SPHERE 


421 


Exercises. Spherical Polygons and Polyhedral Angles 

1. The planes which are perpendicular to the three 
faces of a trihedral angle and bisect the face angles meet 
in a straight line. After proving the proposition state the 
corresponding one relating to a spherical triangle. 

It is unnecessary to prove the latter, since it follows by § 476. 

2. The planes passing through the edges of a trihedral 
angle and perpendicular to the opposite faces meet in a 
straight line. Consider also, as in Ex. 1, the corresponding 
proposition relating to a spherical triangle. 

3. Find the area of a spherical triangle, given that the 
perimeter of its polar triangle is 297° and that the radius 
of the sphere is 10 in. 

4. Find the spherical excess of a spherical triangle 
whose angles are 87°, 92°, and 106°; of a spherical quadri¬ 
lateral whose angles are 145°, 92°, 75°, and 125°. 

5. If the two polygons of Ex. 4 are both on a sphere of 
radius 10 in., what is the area of each ? 

On a sphere of radius 7 in.y find the areas of spherical 
triangles with angles as follows: 

6. 92°, 93°, 94°. 8. 100°, 100°, 100°. 10. 32°, 48°, 130°. 

7. 98°, 102°, 116°. 9. 98°, 102°, 120°. 11. 68°, 37°, 140°. 

On a sphere of radius U in., find the areas of spherical 
polygons with angles as follows: 

12. 80°, 90°, 100°, 110°. 14. 96°, 72°, 116°, 130°. 

13. 72°, 88°, 110°, 120°. 15. 100°, 100°, 100°, 100°. 

16. Discuss the case of the area of a spherical triangle 

whose angles are 200°, 280°, and 60° on a sphere whose 
radius is 10 in. 


422 


GENERAL REVIEW 


BOOK VIII 


Exercises. Miscellaneous Exercises 

1. If a cube and a sphere have equal volumes, what is 
the ratio of the radius of the sphere to the edge of the cube ? 

2. Given that the diagonal of a cube is 4 \/3 in., find the 
radius of the sphere whose area is equal to that of the cube. 

3. The radius of the base of a right circular cylinder is r 
and the altitude of the cylinder is h. Find the radius and 
the volume of a sphere whose area is equivalent to the 
lateral surface of the cylinder. 

4. If the area of a zone of one base is n times the area 
of the circle which forms this base, the altitude of the zone 
is equal to the diameter of the sphere multiplied by (n—l)/n. 
Discuss the special case in which n = l. 

5. Find to the nearest 0.1 in. how far from the center of 
a sphere of radius 8 in. a plane should be passed so as to 
cut from the sphere a circle 154 sq. in. in area. 

6. Find the ratio of the volume of a sphere to the volume 
of the inscribed cube. 

7. Consider Ex. 6 for the circumscribed cube. 

8. Find the ratio of the volume of the cube inscribed in 
a sphere to that of the cube circumscribed about the sphere. 

9. Find the difference between the volumes of two 
cubes, one inscribed in a sphere of radius 1 in. and the 
other circumscribed about it. 

10. Find the difference between the volume of a frustum 
of a pyramid and the volume of a prism each 20 ft. high, 
given that the bases of the frustum are squares 20 ft. and 
12 ft. respectively on a side, and the base of the prism is the 
section of the frustum parallel to the bases and midway 
between them. 

Omit Ex. 10 if § 408 was not taken. 


§513 


MISCELLANEOUS EXERCISES 


423 


11. In certain parts of the United States, stacks of hay 
are shaped roughly like a barn, as shown in this figure. 
The farmers use this rule for finding the approximate num¬ 
ber of tons: Take the "'overthrow'^ (the length ABODE)] 
subtract the width; divide by 2, 
and call the result the height. 

Multiply this height by the prod¬ 
uct of the length and width, and 
call the result the cubic contents. 

Divide this by 412 to find the num¬ 
ber of tons of wild hay; by 450 for timothy; and by 512 
for alfalfa or clover. Consider the accuracy of the rule for 
finding the volume in the case here shown. 

12. Using the rule and the dimensions of the figure in 
Ex. 11, find the approximate number of tons in a stack of 
wild hay; of timothy ; of clover. 

13. The square of the diagonal of a cube is how many 
times the square of an edge ? 

14. Find the ratio of the sum of the squares of all the 
edges of a cube to the sum of the squares of all the diag¬ 
onals of the faces; of all the diagonals of the cube itself. 

15. Find the length of the perpendicular from a vertex 
of a regular octahedron of edge e to the plane determined 
by the four adjacent vertices. 

16. The shortest distance between two opposite edges 
of a regular tetrahedron is equal to half the diagonal of 
the square constructed on an edge. 

17. The sum of the squares of the edges of any tetra¬ 
hedron is four times the sum of the squares of all the lines 
joining the midpoints of opposite sides. 

18. Find the volume of the regular tetrahedron of which 
the sum of the areas of the faces is 4/. 







424 


GENERAL REVIEW 


BOOK VIII 


19. The six planes that pass through the six edges of 
a tetrahedron and bisect the respective opposite edges 
meet in a point. 

20. A cubic foot of copper is drawn into a wire 2000 ft. 
long. Find the diameter of the wire. 

21. Find the volume of a pyramid with equal lateral 
edges e and a base which is an equilateral triangle of side s. 

22. Consider Ex. 21 for the case in which the base is a 
regular octagon of side s. 

23. The base of a regular pyramid of volume V and 
height /i is a square. Find the length of a lateral edge. 

24. In a cube a plane passes through the midpoints of 
three edges that meet in the same vertex. Given that the 
volume of the cube is i, find the volume of the tetrahedron 
thus cut off. 

25. An iron casting is in the form of a right circular cone 
upon the base of which is a hemisphere of the same radius 
outside the cone. If the casting, which is 7 in. in diameter 
and in. long, is placed in a cylindric can 8f in. in diam¬ 
eter and 10 in. high, filled with water, how much water 
remains in the can? 

26. Water is flowing into a tank through a pipe 2.1 in. in 
diameter at the rate of 3 ft. (linear) per second. Allowing 
231 cu. in. to a gallon, how much water will flow into the 
tank in 1 hr. ? 

27. If the centers of two intersecting spheres are 5 in. 
apart, and the radii of the spheres are 3 in. and 4 in. re¬ 
spectively, what is the area of the circle formed by their 
intersection ? 

28. A sphere 1 ft. in diameter is cut from a cube of lead 
1 ft. on an edge. If the pieces cut off are melted and cast 
into another sphere, what is the diameter of this sphere ? 


SUPPLEMENT 


I. Incommensurable Cases 

514. Subjects Treated. In the study of geometry there 
are many topics that might be taken in addition to those 
found in any textbook. The theorems and problems already 
given in this text are standard propositions which are 
looked upon as fundamental, and are usually required as 
preliminary to more advanced work. These propositions and 
a reasonable number of originals selected from the exer¬ 
cises, will be all that most classes have time to consider. 
It occasionally happens, however, that a class is able to do 
more than this, and then more exercises may be selected 
from the large number supplied, and a few additional 
topics may be studied. For this latter purpose the supple¬ 
mentary work is added, but its study should not be under¬ 
taken at the expense of the fundamental propositions and 
exercises. The work on practical mensuration (§§ 543-550), 
however, may be taken with advantage in place of some 
of the less important propositions in the text. 

The subjects treated in the following pages include the 
incommensurable cases of certain propositions, additional 
propositions in the mensuration of solids, a few general 
theorems relating to similar polyhedrons, and some work on 
congruent spherical triangles and on practical mensuration. 
There are also added a few of those recreations of geom¬ 
etry that add a peculiar interest to the subject, and a brief 
sketch of the history of geometry, which all students are 
advised to read as a matter of general information. 

425 


PS 


426 


INCOMMENSURABLE CASES supplement 


515. Central Angles. In § 170 it was proved for the com¬ 
mensurable case that central angles have the same ratio 
as their arcs. We shall now prove the theorem for the 
incommensurable case. 

That is, in the figure here shown in which the AAOB 
and BOC and their arcs AB and BC are incommensurable, 
we have to prove that 

A BOC _ arc BC 
AAOB arcA5 

Divide AAOB into any number of 
equal parts and apply one of these 
parts, as AAOM, to A BOC as many 
times as possible. Since the angles are incommensurable, 
there is a remainder, AXOC, less than one of the parts. 

If we increase the number of parts into which AAOB is 
divided, the size of a part can be decreased indefinitely. 



That is, 
and hence 
Then 
and 
Since 

then 


ZAOM-vO, 

AXOC^O, 
ABOX-^ABOC 
arc BX arc BC. 

AAOB and arc AR are constants, 
ABOX ^ ABOC ^ 

AAOB AAOB' 


and 

But 


slycBX ^ slycBC 

wccAB 2 iYQ,AB 
ABOX _ 2 iYcBX 
AAOB 2iXQ,AB 
ABOC _ 2iYQ,BC 
AAOB arcAR 


§ 301,1 
§170 
§ 301, 2 


That is, the central angles have the same ratio as their 
arcs, even though the angles are incommensurable. 













§§ 515, 516 


SIDES OF A TRIANGLE 


427 


516. Sides of a Triangle. In § 201 it was proved for the 
commensurable case that a line parallel to one side of a 
triangle divides the other sides proportionally. We shall 
now prove the theorem for the 
incommensurable case. 

Divide CD into any number of 
equal parts and apply one of 
these parts, as CM, to DA as 
many times as possible. Since 
CD and DA are incommensurable, there is a remainder, such 
as XA, which is less than one of the parts. 



Construct XY II to AB. § 107 


Then 


DX_EY 
CD CE' 


§201 


If we increase the number of parts into which CD is 
divided, the length of a part can be decreased indefinitely. 


That is. 

CM- 

-^ 0 , 


and hence 

XA- 

-^ 0 , 


and 

YB- 

-> 0 . 


Then 

DX 

-^DA, 


and 

EY 

-^EB, 


Now 

DX 

_^DA 

cd' 


CD 


and 

EY 


§ 301,1 

CE 

ce' 

But 

DX 

EY 

Proved 

CD 

ce' 



. DA 

_EB 

§ 301, 2 


* * CD 

ce' 


That is, the sides are divided proportionally, even though 
their segments are incommensurable. 








428 


INCOMMENSURABLE CASES supplement 


517. Volume of a Rectangular Parallelepiped. Referring to 
the case discussed in § 391, let us suppose that the edges of 
the rectangular parallelepiped are all incommensurable. 

Let Uy in the figure below, be the unit of volume and 
let u be the unit of length. Then if u is applied to AB as 
many times as possible, there is a remainder ri less than u. 
Similarly, there is a remainder 
on the width BE, and a remainder 
rg on the height AD. 

Now if we let the unit U decrease 
indefinitely, and also de¬ 

crease indefinitely; that is, as U^O, 
then u-^0, n—>0, and 

n-^0. 

Since r^, and all approach zero as a limit, we see that 
BZ^l, BW-^w, and BX-^h, as U is taken continually 
smaller and smaller. 

Let P be the volume of the rectangular parallelepiped 
with the dimensions I, w, h, and let P' be the volume of the 
one with the dimensions BZ, BW, BX. 

Now, as C/->0, 

P'-^P. 

From previous discussions of limits we shall assume, as 
seems evidently to be the case, that 
BZ-BW‘BX-^lwh. 

But P' = BZBWBX. §391 

.\P = lwh. §301,2 

No proof of this case is satisfactory for a textbook of this type. If 
rigorous, the proof is too difficult for an elementary class; if simple, 
it lacks scientific accuracy. The fact that the elementary proofs often 
given are open to serious scientific criticism has led most careful writers 
to outline merely the general nature of the proof as has been done 
above. Teachers are advised to require only that the above discussion 
be read understandingly. 















§§ 517, 518 


POLYHEDRONS 


429 


11. Polyhedrons 

Proposition 1. The Polyhedron Theorem 

518. Theorem. In a polyhedron the number of edges 
increased by two is equal to the number of vertices in¬ 
creased by the number of faces. 



Given AG^ a polyhedron; c, the number of edges; v, the 
number of vertices; and /, the number of faces. 

Prove that e-\-2 = v +/. 

Proof. For one face, as BCGF, e = v. 

Adding a second face, as ABCD, there is formed a surface 
of two faces which has one edge (BCX and two vertices 
{B and C), common to the two faces.* 

Hence for two faces, e = v-{-l. 

Adding a third face ABFE, adjoining each of the first 
two, this face will have two edges {AB, BF) and three ver¬ 
tices (A, B, F) in common with the surface of two faces. 

Hence for three faces, e = v-\-2. 

Similarly, for four faces, e = + 3, and so on. 

Hence for (/— 1) faces, e = v +(/— 1)— 1. 

Now the addition of the next face, which is the last one, 
will not increase the number of edges or vertices. 

Hence for / faces, e = v +/— 2, or e + 2 = +/. 







430 


POLYHEDRONS 


SUPPLEMENT 


Proposition 2. Truncated Triangular Prism 

519. Theorem. A truncated triangular prism is equiva¬ 
lent to the sum of three pyramids whose common base is 
the base of the prism and whose vertices are the three 
vertices of the inclined section. 



Given the truncated triangular prism with base ABC and 
inclined section DEFy and divided into the three pyramids 
D-ABCy E-ABCy and F-ABC. 

Prove that ABC-DEF is equivalent to the sum of the three 
pyramids D-ABC, E-ABC, and F-ABC. 

Proof. Dividing the truncated prism ABC-DEF into the 
pyramids E-ACD, E-ABC, E-CFD, as shown in the second 
group of figures, we shall now show that these pyramids 
are equivalent to those in the first group. 












§§ 519-521 


TRUNCATED PRISM 


431 


Now pyramid£7-ACD = pyramidR-ACA §407 

because they have the common base ACD and equal altitudes, since 
the vertices Band B lie on EB which isWto the plane ACD. 

But the pyramid B-ACD may be regarded as having the 
base ABC and the vertex D ; that is, as pyramid D-ABC. 
pyramid E-ACD = pyramid D-ABC. 

The pyramids E-ABC are the same in each division of 
the prism; that is, they have the base ABC and the vertex E. • 

Now A CFD = A CFA, § 245 

because they have the common base CF and equal altitudes, 
since their vertices lie on AD which is II to CF. 

Then pyramid E-CFD = pyramid B-CFA, § 407 

because they have equivalent bases {the A CFD and CFA) and 
equal altitudes, since EB is lUo the plane ACFD. 

But the pyramid B-CFA may be regarded as having the 
base ABC and the vertex F; that is, as pyramid F-ABC. 

pyramid E-CFD = pyramid F-ABC. 

Hence the truncated triangular prism ABC-DEF is equiva¬ 
lent to the sum of the three pyramids whose common base 
is ABC and whose vertices are D, E, and F. 

520. Corollary. The volume of a truncated right triangular 
prism is one third the product of the base 
and the sum of the lateral edges. 

Since the lateral edges DA, EB, FC are ± to 
the base ABC, they are the altitudes of the three 
pyramids whose sum is equivalent to ABC-DEF. 

It is interesting to consider the special case in 
which ADEF is II to A ABC. 

521. Similar Polyhedrons. Polyhedrons which have the 
same number of faces, respectively similar and similarly 
placed, and which have their corresponding polyhedral 
angles equal, are called similar polyhedrons. 






432 


POLYHEDRONS 


SUPPLEMENT 


Proposition 3. Similar Polyhedrons 


522. Theorem. Two similar polyhedrons can he sepa¬ 
rated into the same number of tetrahedrons^ similar each 
to each and similarly placed. 





B' C 


B 


C 


Given the similar polyhedrons P and P'. 

Prove that P and P' can he separated into the same number 
of tetrahedrons^ similar each to each and similarly placed. 

Proof. Let E and E' be corresponding vertices. 

By drawing corresponding diagonals, as AC, A'C\ let all 
the faces of P and P\ except those which include the AE 
and E\ be divided into corresponding A. 

Also, let a plane, as EAC, pass through E and each 
diagonal of the faces of P, and a plane, as E'A'C\ through 
E' and each corresponding diagonal of P'. 

Any two corresponding tetrahedronsP-APC and E'-A'B'C' 
have the faces ABC, EAB, EBC similar respectively to the 


faces A'B’C', E'A’B’, E'B’C\ 


§225 



§205 


face EAC is similar to face E'A'C'. 


then 


§214 










§§ 522-527 SIMILAR POLYHEDRONS 433 

Then the corresponding trihedral A of the tetrahedrons 
are equal. § 367 

Hence E-ABC is similar to E'-A'B'C\ § 521 

If E-ABC and E'-A'B’C’ are removed, the polyhedrons 
which are left remain similar; for the new faces EAC and 
E'A'C' have just been proved similar, the modified faces 
AED and A'E'D\ ECF and E'C'F\ are similar (§225), and 
the modified polyhedral AE and E\ A and A', C and C' 
remain equal each to each, since the corresponding parts 
taken from these A are equal. This process of removing 
similar tetrahedrons can be continued as necessary. 

Hence P and P' can be separated into the same number of 
tetrahedrons, similar each to each and similarly placed. 

523. Corollary. The corresponding edges of similar poly¬ 
hedrons are proportional. 

This follows from the definitions of §§ 521 and 205. 

524. Corollary. Any two corresponding lines in two similar 
polyhedrons have the same ratio as any two corresponding 
edges. 

For these lines may be shown to be sides of similar polygons. 

525. Corollary. Two corresponding faces of similar poly¬ 
hedrons are proportional to the squares of any two corre¬ 
sponding edges. 

This follows from §§ 521 and 251. 

526. Corollary. The areas of the entire surfaces of two 
similar polyhedrons are proportional to the squares of any 
two corresponding edges. 

527. Corollary. The areas of two similar cylinders, or of 
two similar cones, are proportional to the squares of any 
two corresponding lines. 

Consider limits, and apply § 526. 


434 


POLYHEDRONS 


SUPPLEMENT 


Proposition 4. Ratio of Tetrahedrons 

528. Theorem. The volumes of two tetrahedrons which 
have a trihedral angle of one equal to a trihedral angle 
of the other are to each other as the products of the three 
edges of these trihedral angles. 



Given S-ABC and S'-A’B'C'y two tetrahedrons with trihedral 
ZS = trihedral Z.S'; andF andF', the volumes. 


Prove that 


V _ SA-SB-SC 

V S'A’ • S'B' • S'C ’ 


Proof. Place tetrahedron S’-A'B'C' upon S-ABC so that 
trihedral ZS shall coincide with its equal, ZS'. 

Let CD and C'D' be Js to the plane SAB, and let the plane 
of CD and C'D' intersect SAB in SD'D. 

The faces SAB and SA'B' may be taken as the bases and 
CD and C'D' as the altitudes of the triangular pyramids 
C-SAB and C'-SA'B' respectively. 


Then _ SAB CD 

V iSA'B' • C'D' SA'B'' C'D' * 


§406 


But 


SAB 
SA'B' ■ 


= ^(1249), and = 205). 


. V _ SA-SB-SC _ SA^SB^SC 
*‘F' SA'-SB'-SC S'A'-S'B'-S'C' 


Ax. 5 

















§§ 528, 529 


RATIO OF TETRAHEDRONS 


485 


Proposition 5. Ratio of Tetrahedrons 

529. Theorem. The volumes of two similar tetrahedrons 
are to each other as the cubes of any two corresponding 
edges. 



Given P-ABC and P^-A'B'C'y two similar tetrahedrons; V and 
V'f the volumes; and PB and P'5', two corresponding edges. 


Prove that 


F. 

v 


PB 

Fb'^ 


Proof. Since P-ABC is similar to P'-A'B'C', 

the corresponding polyhedral A are equal. 

Hence 


or 


But 


V 

PB 

PC- 

PA 

r 

P'B'‘ 

P'C 

•P'A'' 

V _ 

PB 

PC 

PA 

V’ 

P'B'' 

P'C 

' P'A' 

PB 

- PC __ 

PA 

P'B' 

P'C' P'A' 


Substituting for its equals, we have 
V PB PB PB 


r 


or 


P'B' P'B' P'B' 
V _ pF 

V' Fb'"^ 


Given 

§521 

§528 


§523 


Ax. 5 












436 


POLYHEDRONS 


SUPPLEMENT 


Proposition 6. Ratio of Polyhedrons 

530. Theorem. The volumes of two similar polyhedrons 
are to each other as the cubes of any two corresponding 
edges. 



Given P and P'^ two similar polyhedrons; V and the vol¬ 
umes; and EB and E'B', any two corresponding edges. 

Prove that V:V' = ^’': Wb'\ 

Proof. Let P and P' be separated into tetrahedrons 
similar each to each and similarly placed (§ 522), and let 
their respective volumes be Vi, Fg, Fg, • • Vi Vi Fg, • • •. 

Then : Vi = eS‘ : Wb'^, 

V^:Vi = EB^:E'B'^, and so on. § 529 

.■.V,+V^+V,+ ---:Vi+Vi+Vi+---^W-.WB'\ §198,8 

But Fx+y,+V 3 + • • • = F, and Vi+Vi+Vi+ ■■■=¥'. 

V:V’= ^":Wb'\ Ax. 5 

531. Corollary. The volumes of two similar cylinders^ or 
of two similar cones, are proportional to the cubes of any 
two corresponding lines. 

Consider limits, and apply § 530. 










§§ 530-533 


CAVALIERrS THEOREM 


437 


532. Cavalieri’s Theorem. In connection with the work 
in mensuration it is desirable to call attention to a theo¬ 
rem which was set forth by an Italian mathematician, 
Bonaventura Cavalieri (1598-1647), nearly three centuries 
ago. Since a complete proof requires some knowledge of 
the calculus, the theorem is here treated informally. 


Theorem. If two solids lie between parallel planes^ 
and if sections made by any plane parallel to the given 


planes are equivalent^ the 
solids are equivalent. 

That is, if the two solids 
S and S\ lie between the 
parallel planes m and n, 
and if the planes x, y, • — 
cut the solids S and S' so 



that A = A'jB=B','"y the solids and S' are equivalent. 

Let P and P' be the portions lying between x and 
and let the altitude of P and P' be one nth. of the altitude h 
of S and S', On the bases A, A'; n\ 

By B'] ''' suppose right cylinders 
or prisms C and C' to stand, each 


t 


Jl 


with the altitude h/n. Then C=C', for any value of n. 
As 7^-^oo we see that the sum of the C's approaches S 
and the sum of the C'^s approaches S', Since each C is 
always equivalent to each C', we may assume that <8 = S'. 


533 . Prismoid. A polyhedron which has for its bases two 
polygons in parallel planes, and for its lateral faces triangles 
or trapezoids with one side common to one base and the 
opposite vertex or side common to the other base, is called 
a prismoid. 

The midsection of such a polyhedron is the section which is parallel 
to the base and bisects the altitude. 






















438 


POLYHEDRONS 


SUPPLEMENT 


Proposition 7. Prismoid Formula 

534. Theorem. The volume of a prismoid is the product 
of one sixth the altitude hy the sum of the bases and four 
times the midsection. 



Given CDE • • • -XY • • •, a prismoid; F, the volume; B, and 

ilf, the areas of the bases and midsection; and /i, the altitude. 

Prove that V=^h{B + B'+AM). 

Proof. If any lateral face is a trapezoid, let it be divided 
into two A by a diagonal, as EY. 

Let any point P in the midsection be joined to the ver¬ 
tices of the polyhedron and of the midsection, thus separat¬ 
ing the prismoid into pyramids which have their vertices at 
P, and which have as their respective bases the lower base, 
the upper base, and the lateral faces of the prismoid. 

The pyramid P-XCD, which we may call a lateral pyra¬ 
mid of volume Vp, is composed of the pyramids P-XQR, 
P-QDR, P-QCD of volumes Vj, V^, respectively. 

Now P-XQR may be regarded as having the vertex X 
and base PQR, and P-QDR the vertex D and base PQR. 












§534 


PRISMOID FORMULA 


439 


Then in P-XQR V^=\h' PQR, 
and in P-QDR PQR. § 406 

The pyramids P-QCD and P-QDR have the same vertex 
P, but, since the base CD of AQCD is twice the base QR of 
AQDR (§ 87), and these A have the same altitude (§ 533), 
the base QCD is twice the base QDR (§ 246). 

Hence the pyramid P-QCD is equivalent to twice the 
pyramid P-QDR ; or, in pyramid P-QCD, 

V^=lh-PQR. Ax.3 

Hence in pyramid P-XCD, which is composed of P-XQR, 
P-QDR, and P-QCD, 

Vp=lh‘PQR, Ax.l 

Similarly, the volume of each lateral pyramid is | /i times 
the area of that part of the midsection included within it; 
and hence for the sum of all the lateral pyramids, 

Vp=lh’M. Ax.l 

The volume of the pyramid with base CDE • • • is | hB, and 
that of the pyramid with base XY • • • is ^ hB'. § 406 

.\V=ih{B-^B'+4:M). Ax.l 

Exercises. The Prismoid 

Deduce the following formulas as cases of a prismoid: 

1. Cube, 3. Pyramid, F= ^ P/i. 

2. Prism, U=P/^. 4. Parallelepiped, F= P/i. 

5. Frustum of a pyramid, V=\h{B-\-B'-\-y/BB'). 

6. The area of the upper base of a prismoid is 5 sq. in.; 
of the lower base, 9 sq. in.; of the midsection, 7sq. in.; and 
the altitude is 4 im Find the volume. 

7. Consider Ex. 6 when each measurement is doubled. 


440 


POLYHEDRONS 


SUPPLEMENT 


Exercises. Review 

1. Given that the base of a regular pyramid is an equi¬ 
lateral triangle of side s and that the slant height is I, find 
the altitude and the volume of the pyramid. 

2. Consider Ex. 1 when the base is a square of side s. 

3. Given that the base of a regular pyramid is a square 
of side s and that the area of each lateral face is A, find 
the volume of the pyramid. 

4. Find the volume of a truncated right triangular prism 
whose base has an area of 7 sq. in., and whose lateral edges 
are ijin., l| in., and 2f in. respectively. 

5. In two tetrahedrons which have a trihedral angle of 
one equal to a trihedral angle of the other, the edges of 
this trihedral angle in the first are 3 in., 4 in., and 5 in. 
respectively, and those of the corresponding angle of the 
other are 5 in., 6 in., and 7 in. respectively. Find the ratio 
of the volumes of the tetrahedrons. 

6. How many faces are there in a crystal which has 
five vertices and nine edges ? 

7. What part of a cube is cut off by a plane’passing 
through the vertex B' in the upper base and the diagonal AC 
in the lower base ? 

8. Two similar polyhedrons have the edges e, and of 
the first corresponding to e[ and e' of the second. If = 4 in., 
^ 2 =7 in., and e[ = 5.6 in., how long is e' ? 

9. By the aid of Cavalieri^s Theorem, prove § 405. 

10. A wedge has for its base a rectangle I inches long 
and w inches wide. The cutting edge is e inches long, and 
is parallel to the base. The distance from e to the base is 
d inches. Write a formula for the volume of the wedge. 
Apply this formula to the case of Z = 5, 21 ; = 2, e = 3, d = 4. 


§535 


SPHERICAL TRIANGLES 


441 


III. Spherical Triangles 
Proposition 8. Two Sides and Included Angle 

535. Theorem. If two triangles on the same sphere or 
on equal spheres have two sides and the included angle 
of one equal respectively to the corresponding parts of the 
other, the triangles are either congruent or symmetric. 



Given ABC and two A on the same sphere or on 

equal spheres, with AB = A'B', AC=A'C', Z.A = AA\ and 
similarly arranged; and the A ABC and A'B'X with AB — A^B\ 
AC = A'Xy AA = AA', and arranged in reverse order. 

Prove that A ABC is congruent to AA'B'C\ 
and that A ABC is symmetric with respect to AA'B'X, 

Proof. Place A ABC upon A A'B'C'. Post. 5 

By a proof similar to that of the corresponding case in plane geom- 
try (§ 40), show that AABC is congruent to AA'B'C'. 

Let A A'B'C' be symmetric with respect to AA'B'X Now show that 
AA'B'X and A'B'C' have A'B'=A'B', A'X= A'C', ZXA'B'= Z C'A'B', 
and are arranged in reverse order (§ 487), and hence that AABC and 
A'B'C' have AB = A'B', AC = A'C', ZA = ZC'A'B'(Ax. 5), and are 
similarly arranged. Then show that A ABC and A'B'C' are congruent, 
as above, and hence that AABC and A'B'X are symmetric (Ax. 5). 

PS 














































442 


SPHERICAL TRIANGLES 


SUPPLEMENT 


Proposition 9. Two Angles and Included Side 

536. Theorem. If two triangles on the same sphere or 
on equal spheres have two angles and the included side 
of one equal respectively to the corresponding parts of the 
otherj the triangles are either congruent or symmetric. 







Given ABC and A'R'C', two A on the same sphere or on equal 
spheres, with ZA = ZA', ZC=ZC', AC=A'C', and simi¬ 
larly arranged; and the A ARC and A'B'X with ZA = ZA', 
ZC=ZA', AC = A'X, and arranged in reverse order. 

Prove that A ABC is congruent to AA'B'C'y 

and that A ABC is symmetric with respect to A A'B 'X. 

Proof. Place AARC upon AA'R'C'. Post. 5 

By a proof similar to the corresponding case in plane geometry (§ 44), 
show that AABC is congruent to AA'B'C'. 

Let AA'B'C' he symmetric with respect to AA'B'X. 

Now show that AA'BX and A'B'C' have ZXA'B' = ZC A B', 
ZX = ZC\ AX = A'C', and are arranged in reverse order (§ 487), and 
hence that AABC and A'B'C' have AA = ZC'A'B', ZC = ZC', and 
AC = A'C' (Ax. 5), and are similarly arranged. Then show that AABC 
and A'B'C' are congruent, as proved above, and hence that AARC is 
symmetric with respect to AA'B'X (Ax. 5). 































§§ 536,537 


EQUILATERAL TRIANGLES 


443 


Proposition 10. Mutually Equilateral Triangles 

537. Theorem. If two triangles on the same sphere or 
on equal spheres are mutually equilateral, they are mutu¬ 
ally equiangular and are either congruent or symmetric. 



Given ABC, two A on the same sphere or on equal 

spheres,'with. AB = A^B\ BC = B'C% and CA = C^A\ 

Prove that ZA = ZA', ZC = ZC', and that 

A ABC and A'B'C' are either congruent or symmetric. 

Proof. Let O and O' be the centers of the spheres, and let 
a plane pass through each pair of vertices of each A and 
the center of its sphere. 

Then in the trihedral Z at O and O' the corresponding 
face A are respectively equal. § 137 

Then the trihedral Z O and O' are equal (§ 367), and hence 
the corresponding dihedral Z are respectively equal (§ 364). 

ZA = ZA', AB = AB', AC=AC', §473 

Hence the A are either congruent or symmetric. § 536 

In the above figure the parts are arranged in the same order, so that 
the triangles are congruent. The parts might be arranged in reverse 
order, as in the A ARC and A'B'X in the figure of § 536, in which case 
the A ARC and A'B'C' would be symmetric. 




















444 


SPHERICAL TRIANGLES 


SUPPLEMENT 


Proposition 11. Mutually Equiangular Triangles 

538. Theorem. If two triangles on the same sphere or on 
equal spheres are mutually equiangular, they are mutu¬ 
ally equilateral, and are either congruent or symmetric. 



Given T and T', two mutually equiangular A on the same 
sphere or on equal spheres. 

Prove that T and T' are mutually equilateral, and that 
they are either congruent or symmetric. 

Proof. Let AP be the polar A of AT, and let AP' be the 
polar A of AT'. 

Since AT and T' are mutually equiangular, Given 
the polar A P and P' are mutually equilateral. § 483 

Hence the polar APandP'are mutually equiangular. § 537 

Now A T and T'are the polar A of A P and P'. §482 

Hence AT and T' are mutually equilateral. § 483 
.‘.AT and T' are either congruent or symmetric. § 537 

The statement that mutually equiangular spherical triangles are 
mutually equilateral, and are either congruent or symmetric, is true 
only when they are on the same sphere or on equal spheres. When 
the spheres are unequal, the spherical triangles are unequal. 


























§§ 538, 539 


ISOSCELES TRIANGLE 


445 


Proposition 12. Isosceles Triangle 

539. Theorem. In an isosceles spherical triangle the 
angles opposite the equal sides are equal; and conversely, 
if two angles of a spherical triangle are equal, the sides 
opposite these angles are equal. 



1. Given the isosceles spherical A ABC with AC=:AB. 
Prove that Z.B =/I C. 

Proof. Let AD be the arc of a great O which bisects Z.A. 


Then AB — AC, Given 

AD=AD, Iden. 

and ZBAD=Z CAD. Const. 

Hence ABDA and CD A are symmetric. § 535 

.\Z.B=^Z.C. §487 

2. Given the spherical AABC with = Z.C. 

Prove that AC = AB. 

Proof. Let AA'B'C' be the polar A of AABC. 

Then A'C'+Z5 = 180° and A= 180°. § 483 

.*. A'C' = A'B\ Axs. 5, 2 

ZB' = ZC’. 

AC = AB. 


Then, by 1, 


§ 483, Ax. 5 








446 


SPHERICAL TRIANGLES 


SUPPLEMENT 


Proposition 13. Unequal Parts 

540. Theorem. If two angles of a spherical triangle 
are unequal, the sides opposite these angles are unequal, 
and the side opposite the greater angle is the greater; 
and conversely, if two sides are unequal, the angles oppo¬ 
site these sides are unequal, and the angle opposite the 
greater side is the greater. 



1. Given the spherical /\ABC with Z.OZ.B. 

Prove that AB^AC. 

Proof. Let CD, the arc of a great O, make ADCB = Z.B. 
Then DB = DC. §539,2 

Now AD-\-DC>AC. §479 

.*. AD-\-DB>AC, or AB> AC. Ax. 5 

2. Given the spherical A ABC with AB>AC. 

Prove that /.C>Z.B. 

Proof. Using the indirect method, if ZC = Z5, then 
AB = AC(^ 539, 2), which is impossible, since AB>AC', and 
if ZC<Z5, then, by 1, AB<AC, which is also impossible. 
.*. ZOZB. 












§540 


REVIEW EXERCISES 


447 


Exercises. Review 

1. P'ind the volume of a truncated right triangular prism 
whose lateral edges are 1 in., 1| in., and 2^ in. respectively, 
and whose base is an equilateral triangle 3 sq. in. in area. 

2. The volume of any truncated triangular prism is one 
third the product of a right section and the sum of the 
lateral edges. 

3. In two tetrahedrons which have a trihedral angle of 
one equal to a trihedral angle of the other, the edges of 
these angles are 2 in., 2j in., 3 in. in the first tetrahedron 
and 3 in., 3^- in., ij in. in the second. Find the ratio of the 
volumes of the tetrahedrons. 

4. A polyhedron of eight faces and six vertices has how 
many edges ? 

5. Consider the possibility of a crystal with four faces 
and two edges; with six faces and four edges. 

6. Consider the possibility of a four-edged polyhedron. 

7. The volume of the first of two similar tetrahedrons 
is 32 cu. in., and to an edge 2 in. long in the first there 
corresponds an edge 2|in. long in the second. Find the 
volume of the second tetrahedron. 

8. Given that the volumes of two similar polyhedrons 
are in the ratio 1: 8, find the length of the edge of the 
first that corresponds to one 3 in. long in the second. 

9. Find the volume of a prismoid in which the areas of 
the bases are 7 sq. in. and 4 sq. in. respectively, the area 
of the midsection is 5 sq. in., and the height is 8 in. 

10. Show that the formula for the volume of a prismoid 
also applies to the cylinder and the cone. 

11. Sketch the figure and state the proposition relating 
to trihedral angles that follows from each of §§ 535-539. 


448 


SPHERICAL TRIANGLES 


SUPPLEMENT 


541. spherical Segment. If a sphere is cut by two parallel 
planes, the solid thus formed between the planes is called 
a spherical segment. 

The parallel sections are called the bases. 

The formula for the volume F of a segment is 
V=i7rhi3r^+Sr'^+h^), 

where r and r' are the radii of the bases and h is 
the altitude of the segment. This formula is given 
here only for reference, as its proof is too difficult. 

If one of the parallel planes is tangent to the sphere, the segment 
is called a spherical segment of one base. 



542. Spherical Sector. The solid generated by the revo¬ 
lution of a sector of a circle about a diameter of the circle 
as an axis is called a spherical sector. 

In this figure the solid is generated by the revo¬ 
lution of the sector AOB about the diameter PP'. 

The zone generated by the arc AB is called the base 
of the spherical sector. 

The formula for the volume V of such a solid is 
V=lrS, 

where S is the area of the base, and r is the radius of the sphere. 



Exercises. Spherical Segments and Sectors 

1. Find the volume of a spherical segment whose bases 
are 6 in. and 8 in. in diameter, and whose altitude is 2 in. 

2. If the diameter of a sphere is 14 in. and the altitude 
of the zone forming the base of a spherical sector is 3 in., 
what is the volume of the sector ? 

3. By regarding a spherical segment of one base as the 
difference between a spherical sector (whose base is a zone 
of one base) and a cone, show that the formula for the 
volume \s>V=\irh^(2>r — h), where r is the radius of the 
sphere and h is the altitude of the segment. 






§§ 541-543 


PRACTICAL MENSURATION 


449 


IV. Practical Mensuration 

543. Nature of the Work. In this country the demand for 
supplementary exercises in practical mensuration is in¬ 
creasing. The work which is required is based not merely 
upon the demonstrative geometry of the plane and of the 
simpler solids as set forth in this text, but also upon the 
actual measurements of lines and upon the trigonometry 
of the right triangle as it is now taught in connection with 
algebra in many schools. Material of this kind may safely 
replace certain of the propositions of plane geometry, such 
as those which involve inequalities, certain theorems in rela¬ 
tion to the circle, and certain parts of Book IV, and it may 
also take the place of various propositions in Book VI. It is 
here offered as optional work for the use of those teachers 
who wish to modify the standard course in demonstrative 
geometry, as given in this text, by the introduction of a 
moderate amount of work in advanced mensuration. 

No attempt has been made to include any exercises on 
the mensuration of the conic sections, which is more advan¬ 
tageously treated in the calculus or in connection with the 
propositions of analytic geometry. Furthermore, such work 
is not so practical for the general student as that which 
relates to the more common plane and solid figures. 

As giving proper training in space perception, without 
involving the logic of demonstration, it is believed that 
teachers will find this material of great value. It is gener¬ 
ally conceded that plane geometry furnishes a sufficient 
amount of training in deductive reasoning for an initial 
course, and that the value of solid geometry lies chiefly in 
its presentation of spatial relations. Such a presentation 
is made more vital by work of the nature and extent set 
forth in the following pages. 


450 


PRACTICAL MENSURATION supplement 


544. Symbols and Formulas of Plane Geometry. The fol¬ 
lowing symbols and formulas are needed in the exercises: 

Symbols 

a = apothem A = area r = radius 

b = base C = circumference s = side 

d = diameter Z A = angle of polygon tt = 3.1416 

h = altitude ZO = central angle I/tt = 0.3183 

Primes indicate that there are two parts of the same name, such 
as the bases h and h' of a trapezoid. 

Formulas to be Memorized 
Parallelogram, A = bh Trapezoid, A = ^h(b-\-b') 

Triangle, A — \bh Circumference, C = 2 7rr = ttc^ 

Area of a circle, A = = \ ird^ 

Formulas for Reference 
Chord AB, Cj = 2 r sin J 0 

Chord ACy C 2 = 2 r sin | ^ 

Arc ACB, i (8 C 2 — Cl) ^ 

The result by this formula is approximate. 

Sector OACBy A = irr^d 

Equilateral triangle, A = 0.4330 
ZA = 60° ZO = 120° r = 0.5774 s 

Regular pentagon, A = 1.7205 
ZA = 108° ZO = 72° r = 0.8506 s 
Regular hexagon, A = 2.5980 
ZA = 120° ZO = 60° r = s 
Regular octagon, A = 4.8284 
ZA = 135° ZO = 45° r = 1.3065 s 



a = 0.2887 s 
a = 0.6882 s 

a = 0.8660 s 
a = 1.2071 s 





§544 


FORMULAS 


451 


Simpson's Rule for Areas 

To find the area between the x axis and a continuous 
curve, Simpson^s Rule^ which is 
named for its inventor, is often used. 

In the figure here shown the base 
line AB {x axis) is divided into an 
even number of equal parts so that 
there is an odd number of ordinates 

Viy 2 / 2 » Vzy- "f Vn. The approximate area is then found by 
the following formula: 

^ = J^[(2/1 4 " 2 /w)+2(2/3+2/54-2/7+ * * ■) +4(2/2 + 2/4 + 2 / 6 + * * *)] 

This formula may be expressed in words as follows: 

To find the approximate area between a continuous curve 
and the x axis, add the extreme ordinates, twice the sum of 
the other odd ordinates, and four times the sum of the even 
ordinates, and then multiply the result by one third the com¬ 
mon distance between the successive ordinates. 

The approximation is closer if the curve is undulating (wave-like) 
as in the figure shown above. The formula and rule need not be 
memorized. 

In the work in practical mensuration the symbols (') and (") are 
used for feet and inches respectively. 

The application of the rule may be illustrated by the 
case of a curve which has the ordinates 16.5', 21', 24', 25', 
29.5', 33', 30', 28.5', 28.5', 29', 30', 25.5', 21.5', 20', 20.5', and a 
common distance {h) between the ordinates of 7.68'. 

We then have 2/1 + 2/n = 16.5' -f 20.5' = 37'. 

2 ( 2/3 -f 2/5 + 2/7 +•••) = 2 (24' + 29.5' + 30' + 28.5' + 30' + 21.5') 

= 327'. 

4 ( 2/2 + 2/4 + 2/6 +•••)= 4 (21' + 25' 4 33' + 28.5' + 29' + 25.5' -|- 20') 

= 728'. 

Hence A = ^ X 7.68 (37 -f- 327 + 728) sq. ft. 

= 2795.52 sq. ft. 











452 


PRACTICAL MENSURATION supplement 


545. Symbols and Formulas of Solid Geometry. In addition 
to the symbols and formulas given in § 544, the following 
are needed in the exercises: 

Symbols 

V = volume S = area of curve surface 

B = area of base I = slant height 

E = spherical excess of a polygon 
In the case of cylinders and cones, only cylinders and cones of revo¬ 
lution are considered. 


Formulas to be Memorized 


Cone, 

S=Trrl 

V=i-irr^h 

Cylinder, 

S = 2 Trrh 

V^'Kr% 

Sphere, 

S = 47rr" 

F=|7rr’ 

Zone, 

S = 2Trrh 


Spherical polygon. 

'^tlsETrr^ 


Prism or cylinder. 


V^Bh 

Pyramid or cone. 


V=\Bh 


It is unnecessary to give formulas for the lateral area of a prism 
and for the area of the total surface of a cone or a cylinder, as these 
areas can be found by taking the sum of other known areas. 

Formulas for Reference 
Frustum of a pyramid, V = \ h{BB'BB') 

Frustum of a cone, S = ttZ (r + r') 

V==lh{B+B'+yfm') = lirh(r‘+r'^-\-rr') 
Spherical sector, V = ^ rS 

Here S' is the area of the zone forming the base of the sector. 

Spherical segment of one base, U = i 'n-h^ (3 r — /t) 

Instead of using this formula the student may consider the spherical 
segment as the difference between a spherical sector and a cone. 



§§545-548 TRIGONOMETRIC FORMULAS 453 



546. Trigonometry Presupposed. The exercises assume a 
knowledge of four functions of an 
angle and the ability to use these func¬ 
tions in solving the right triangle. 

The four functions which are 
needed are as follows: 

/i opposite side a ^ • 

sm A = -- 7 -= -; whence a = c sin A 

hypotenuse c 

COSA = adjacent side ^6. ^ ^ ^ 

hypotenuse c 

tan A = opposite side ^ a . « = 5 tan A 

adjacent side b 

cot A = adjacent side ^ b ^ ^ ^ 

opposite side a 

Tables of these functions are given on pages 454-461. 

547. Trigonometric Formulas. The following formulas are 
inserted for reference: 


sin A + cos^ A = 
sin A 


tan A = - 


1 


cos A cot A 


sin A = Vl— cos^A 


sin A tan A 


548. Use of the Tables. In the tables on pages 454-461 the 
functions are given for every 0.1°, or for every 6'. In the 
columns of differences the difference for every 1' is given. 
For example, to find sin 55° 20', find sin 55° 18' on page 455, 
and to it add 3 (for 0.0003) found under 2'; that is, 

sin 55° 20'= 0.8221-f 0.0003 = 0.8224 

In finding cos 35° 45', for example, since the cosine de¬ 
creases as the angle increases, we subtract the difference 
(indicated in the table as a minus difference). Hence 
cos 35° 45'= 0.8121-0.0005 = 0.8116 













454 


NATURAL SINES. 0°-45° 


SUPPLEMENT 


0 

0 . 0 ° 

0 . 1 ° 

IS 

0 . 3 ° 


0 . 5 ° 

0 . 6 ° 

0 . 7 ° 

s 

0 . 9 ° 

+Differences 


0 ' 

6 ' 

12 ' 

18 ' 

24 ' 

30 ' 

36 ' 

42 ' 

48 ' 

64 ' 

1 ' 

2 ' 

3' 

4' 

5' 

0 

.0000 

.0017 

.0035 

.0052 

.0070 

.0087 

.0105 

.0122 

.0140 

.0157 

3 

6 

9 

12 

TF 

1 

.0175 

.0192 

.0209 

.0227 

.0244 

.0262 

.0279 

.0297 

.0314 

.0332 

3 

6 

9 

12 

15 

2 

.0349 

.0366 

.0384 

.0401 

.0419 

.0436 

.0454 

.0471 

.0488 

.0506 

3 

6 

9 

12 

15 

3 

.0523 

.0541 

.0558 

.0576 

.0593 

.0610 

.0628 

.0645 

.0663 

.0680 

3 

6 

9 

12 

15 

4 

.0698 

.0715 

.0732 

.0750 

.0767 

.0785 

.0802 

.0819 

.0837 

.0854 

3 

6 

9 

12 

14 

5 

.0872 

.0889 

.0906 

.0924 

.0941 

.0958 

.0976 

.0993 

.1011 

.1028 

3 

6 

9 

12 

14 

6 

.1045 

.1063 

.1080 

.1097 

.1115 

.1132 

.1149 

.1167 

.1184 

.1201 

3 

6 

9 

12 

14 

7 

.1219 

.1236 

.1253 

.1271 

.1288 

.1305 

.1323 

.1340 

.1357 

.1374 

3 

6 

9 

12 

14 

8 

.1392 

.1409 

.1426 

.1444 

.1461 

.1478 

.1495 

.1513 

.1530 

.1547 

3 

6 

9 

12 

14 

9 

.1564 

.1582 

.1599 

.1616 

.1633 

.1650 

.1668 

.1685 

.1702 

.1719 

3 

6 

9 

12 

14 

10 

.1736 

.1754 

.1771 

.1788 

.1805 

.1822 

.1840 

.1857 

.1874 

.1891 

3 

6 

9 

11 

14 

11 

.1908 

.1925 

.1942 

.1959 

.1977 

.1994 

.2011 

.2028 

.2045 

.2062 

3 

6 

9 

11 

14 

12 

.2079 

.2096 

.2113 

.2130 

.2147 

.2164 

.2181 

.2198 

.2215 

.2233 

3 

6 

9 

11 

14 

13 

.2250 

.2267 

.2284 

.2300 

.2317 

.2334 

.2351 

.2368 

.2385 

.2402 

3 

6 

8 

11 

14 

14 

.2419 

.2436 

.2453 

.2470 

.2487 

.2504 

.2521 

.2538 

.2554 

.2571 

3 

6 

8 

11 

14 

16 

.2588 

.2605 

.2622 

.2639 

.2656 

.2672 

.2689 

.2706 

.2723 

.2740 

3 

6 

8 

11 

14 

16 

.2756 

.2773 

.2790 

.2807 

.2823 

.2840 

.2857 

.2874 

.2890 

.2907 

3 

6 

8 

11 

14 

17 

.2924 

.2940 

.2957 

.2974 

.2990 

.3007 

.3024 

.3040 

.3057 

.3074 

3 

6 

8 

11 

14 

18 

.3090 

.3107 

.3123 

.3140 

.3156 

.3173 

.3190 

.3206 

.3223 

.3239 

3 

6 

8 

11 

14 

19 

.3256 

.3272 

.3289 

.3305 

.3322 

.3338 

.3355 

.3371 

.3387 

.3404 

3 

5 

8 

11 

14 

20 

.3420 

.3437 

.3453 

.3469 

.3486 

.3502 

.3518 

.3535 

.3551 

.3567 

3 

5 

8 

11 

14 

21 

.3584 

.3600 

.3616 

.3633 

.3649 

.3665 

.3681 

.3697 

.3714 

.3730 

3 

5 

8 

11 

14 

22 

.3746 

.3762 

.3778 

.3795 

.3811 

.3827 

.3843 

.3859 

.3875 

.3891 

3 

5 

8 

11 

14 

23 

.3907 

.3923 

.3939 

.3955 

.3971 

.3987 

.4003 

.4019 

.4035 

.4051 

3 

5 

8 

11 

14 

24 

.4067 

.4083 

.4099 

.4115 

.4131 

.4147 

.4163 

.4179 

.4195 

.4210 

3 

5 

8 

11 

13 

25 

.4226 

.4242 

.4258 

.4274 

.4289 

.4305 

.4321 

.4337 

.4352 

.4368 

3 

5 

8 

11 

13 

26 

.4384 

.4399 

.4415 

.4431 

.4446 

.4462 

.4478 

.4493 

.4509 

.4524 

3 

5 

8 

10 

13 

27 

.4540 

.4555 

.4571 

.4586 

.4602 

.4617 

.4633 

.4648 

.4664 

.4679 

3 

5 

8 

10 

13 

28 

.4695 

.4710 

.4726 

.4741 

.4756 

.4772 

.4787 

.4802 

.4818 

.4833 

3 

5 

8 

10 

13 

29 

.4848 

.4863 

.4879 

.4894 

.4909 

.4924 

.4939 

.4955 

.4970 

.4985 

3 

5 

8 

10 

13 

30 

.5000 

.5015 

.5030 

.5045 

.5060 

.5075 

.5090 

.5105 

.5120 

.5135 

3 

5 

8 

10 

13 

31 

.5150 

.5165 

.5180 

.5195 

.5210 

.5225 

.5240 

.5255 

.5270 

.5284 

2 

5 

7 

10 

12 

32 

.5299 

.5314 

.5329 

.5344 

.5358 

.5373 

.5388 

.5402 

.5417 

.5432 

2 

5 

7 

10 

12 

33 

.5446 

.5461 

.5476 

.5490 

.5505 

.5519 

.5534 

.5548 

.5563 

.5577 

2 

5 

7 

10 

12 

34 

.5592 

.5606 

.5621 

.5635 

.5650 

.5664 

.5678 

.5693 

.5707 

.5721 

2 

5 

7 

10 

12 

36 

.5736 

.5750 

.5764 

.5779 

.5793 

.5807 

.5821 

.5835 

.5850 

.5864 

2 

5 

7 

9 

12 

36 

.5878 

.5892 

.5906 

.5920 

.5934 

.5948 

.5962 

.5976 

.5990 

.6004 

2 

5 

7 

9 

12 

37 

.6018 

.6032 

.6046 

.6060 

.6074 

.6088 

.6101 

.6115 

.6129 

.6143 

2 

5 

7 

9 

12 

38 

.6157 

.6170 

.6184 

.6198 

.6211 

.6225 

.6239 

.6252 

.6266 

.6280 

2 

5 

7 

9 

11 

39 

.6293 

.6307 

.6320 

.6334 

.6347 

.6361 

.6374 

.6388 

.6401 

.6414 

2 

4 

7 

9 

11 

40 

.6428 

.6441 

.6455 

.6468 

.6481 

.6494 

.6508 

.6521 

.6534 

.6547 

2 

4 

7 

9 

11 

41 

.6561 

.6574 

.6587 

.6600 

.6613 

.6626 

.6639 

.6652 

.6665 

.6678 

2 

4 

7 

9 

11 

42 

.6691 

.6704 

.6717 

.6730 

.6743 

.6756 

.6769 

.6782 

.6794 

.6807 

2 

4 

6 

9 

11 

43 

.6820 

.6833 

.6845 

.6858 

.6871 

.6884 

.6896 

.6909 

.6921 

.6934 

2 

4 

6 

8 

11 

44 

.6947 

.6959 

.6972 

.6984 

.6997 

.7009 

.7022 

.7034 

.7046 

.7059 

2 

4 

6 

8 

10 


All the above sines are less than 1, 



































§548 


NATURAL SINES. 45"-90^ 


455 


o 

0.0° 

0.1° 


o 

CO 

O 

O 

o 

0.6° 

0.6° 

o 

o 


0.9° 

-{-Differences 


0' 

6' 

12' 

18' 

24' 

b 

CO 

36' 

42' 

00 

64' 

1' 

2' 

3' 

4' 

5' 

45 

.7071 

.7083 

.7096 

.7108 

.7120 

.7133 

.7145 

.7157 

.7169 

.7181 

2 

4 

6 

8 

10 

46 

.7193 

.7206 

.7218 

.7230 

.7242 

.7254 

.7266 

.7278 

.7290 

.7302 

2 

4 

6 

8 

10 

47 

.7314 

.7325 

.7337 

.7349 

.7361 

.7373 

.7385 

.7396 

.7408 

.7420 

2 

4 

6 

8 

10 

48 

.7431 

.7443 

.7455 

.7466 

.7478 

.7490 

.7501 

.7513 

.7524 

.7536 

2 

4 

6 

8 

10 

49 

.7547 

.7559 

.7570 

.7581 

.7593 

.7604 

,7615 

.7627 

.7638 

.7649 

2 

4 

6 

8 

9 

50 

.7660 

.7672 

.7683 

.7694 

.7705 

.7716 

.7727 

.7738 

.7749 

.7760 

2 

4 

6 

7 

9 

51 

.7771 

.7782 

.7793 

.7804 

.7815 

.7826 

.7837 

.7848 

.7859 

.7869 

2 

4 

5 

7 

9 

52 

.7880 

.7891 

.7902 

.7912 

.7923 

.7934 

.7944 

.7955 

.7965 

.7976 

2 

4 

5 

7 

9 

53 

.7986 

.7997 

.8007 

.8018 

.8028 

.8039 

.8049 

.8059 

.8070 

.8080 

2 

3 

5 

7 

9 

54 

.8090 

.8100 

.8111 

.8121 

.8131 

.8141 

.8151 

.8161 

.8171 

.8181 

2 

3 

5 

7 

8 

55 

.8192 

.8202 

.8211 

.8221 

.8231 

.8241 

.8251 

.8261 

.8271 

.8281 

2 

3 

5 

7 

8 

56 

.8290 

.8300 

.8310 

.8320 

.8329 

.8339 

.8348 

.8358 

.8368 

.8377 

2 

3 

5 

6 

8 

57 

.8387 

.8396 

.8406 

.8415 

.8425 

.8434 

.8443 

.8453 

.8462 

.8471 

2 

3 

5 

6 

8 

58 

.8480 

.8490 

.8499 

.8508 

.8517 

.8526 

.8536 

.8545 

.8554 

.8563 

2 

3 

5 

6 

8 

59 

.8572 

.8581 

.8590 

.8599 

.8607 

.8616 

.8625 

.8634 

,8643 

.8652 

1 

3 

4 

6 

7 

60 

.8660 

.8669 

.8678 

.8686 

.8695 

.8704 

.8712 

.8721 

.8729 

.8738 

1 

3 

4 

6 

7 

61 

.8746 

.8755 

.8763 

.8771 

.8780 

.8788 

.8796 

.8805 

.8813 

.8821 

1 

3 

4 

6 

7 

62 

.8829 

.8838 

.8846 

.8854 

.8862 

.8870 

.8878 

.8886 

.8894 

.8902 

1 

3 

4 

5 

7 

63 

.8910 

.8918 

.8926 

.8934 

.8942 

.8949 

.8957 

.8965 

.8973 

.8980 

1 

3 

4 

5 

6 

64 

.8988 

.8996 

.9003 

.9011 

.9018 

.9026 

.9033 

.9041 

.9048 

.9056 

1 

3 

4 

5 

6 

65 

.9063 

.9070 

.9078 

.9085 

.9092 

.9100 

.9107 

.9114 

.9121 

.9128 

1 

2 

4 

5 

6 

66 

.9135 

.9143 

.9150 

.9157 

.9164 

.9171 

.9178 

.9184 

.9191 

.9198 

1 

2 

3 

5 

6 

67 

.9205 

.9212 

.9219 

.9225 

.9232 

.9239 

.9245 

.9252 

.9259 

.9265 

1 

2 

3 

4 

6 

68 

.9272 

.9278 

.9285 

.9291 

.9298 

.9304 

.9311 

.9317 

.9323 

.9330 

1 

2 

3 

4 

5 

69 

.9336 

.9342 

.9348 

.9354 

.9361 

.9367 

.9373 

.9379 

.9385 

.9391 

1 

2 

3 

4 

5 

70 

.9397 

.9403 

.9409 

.9415 

.9421 

.9426 

.9432 

.9438 

.9444 

.9449 

1 

2 

3 

4 

5 

71 

.9455 

.9461 

.9466 

.9472 

.9478 

.9483 

.9489 

.9494 

.9500 

.9505 

1 

2 

3 

4 

5 

72 

.9511 

.9516 

.9521 

.9527 

.9532 

.9537 

.9542 

.9548 

.9553 

.9558 

1 

2 

3 

4 

4 

73 

.9563 

.9568 

.9573 

.9578 

.9583 

.9588 

.9593 

.9598 

.9603 

.9608 

1 

2 

2 

3 

4 

74 

.9613 

.9617 

.9622 

.9627 

.9632 

.9636 

.9641 

.9646 

.9650 

.9655 

1 

2 

2 

3 

4 

75 

.9659 

.9664 

.9668 

.9673 

.9677 

.9681 

.9686 

.9690 

.9694 

.9699 

1 

1 

2 

3 

4 

76 

.9703 

.9707 

.9711 

.9715 

.9720 

.9724 

.9728 

.9732 

.9736 

.9740 

1 

1 

2 

3 

3 

77 

.9744 

.9748 

.9751 

.9755 

.9759 

.9763 

.9767 

.9770 

.9774 

.9778 

1 

1 

2 

3 

3 

78 

.9781 

.9785 

.9789 

.9792 

.9796 

.9799 

.9803 

.9806 

.9810 

.9813 

1 

1 

2 

2 

3 

79 

.9816 

.9820 

.9823 

.9826 

.9829 

.9833 

.9836 

.9839 

.9842 

.9845 

1 

1 

2 

2 

3 

80 

.9848 

.9851 

.9854 

.9857 

.9860 

.9863 

.9866 

.9869 

.9871 

.9874 

0 

1 

1 

2 

2 

81 

.9877 

.9880 

.9882 

.9885 

.9888 

.9890 

.9893 

.9895 

.9898 

.9900 

0 

1 

1 

2 

2 

82 

.9903 

.9905 

.9907 

.9910 

.9912 

.9914 

.9917 

.9919 

.9921 

.9923 

0 

1 

1 

2 

2 

83 

.9925 

.9928 

.9930 

.9932 

.9934 

.9936 

.9938 

.9940 

.9942 

.9943 

0 

1 

1 

1 

2 

84 

.9945 

.9947 

.9949 

.9951 

.9952 

.9954 

.9956 

.9957 

.9959 

.9960 

0 

1 

1 

1 

1 

85 

.9962 

.9963 

.9965 

.9966 

.9968 

.9969 

.9971 

.9972 

.9973 

.9974 

0 

0 

1 

1 

1 

86 

.9976 

.9977 

.9978 

.9979 

.9980 

.9981 

.9982 

.9983 

.9984 

.9985 

0 

0 

1 

1 

1 

87 

.9986 

.9987 

.9988 

.9989 

.9990 

.9990 

.9991 

.9992 

.9993 

.9993 

0 

0 

0 

1 

1 

88 

.9994 

.9995 

.9995 

.9996 

.9996 

.9997 

.9997 

.9997 

.9998 

.9998 

0 

0 

0 

0 

0 

89 

.9998 

.9999 

.9999 

.9999 

.9999 

1.000 

1.000 

1.000 

1.000 

1.000 

0 

0 

0 

0 

0 


The precise value of all sines except sin 90° is less than 1. 






























456 


NATURAL COSINES. 0"-45' 


SUPPLEMENT 


o 

0.0° 

0.1° 

2 

2 


0.5° 

0.6° 

0.7° 

0.8° 

0.9° 

•— Differences 


0' 

6' 

12' 

18' 

24' 

30' 

36' 

42' 

00 

54' 

1' 

r 

3' 

4' 

5' 

0 

1.000 

1.000 

1.000 

1.000 

1.000 

1.000 

.9999 

.9999 

.9999 

.9999 

0 

0 

0 

0 

0 

1 

.9998 

.9998 

.9998 

.9997 

.9997 

.9997 

.9996 

.9996 

.9995 

.9995 

0 

0 

0 

0 

0 

2 

.9994 

.9993 

.9993 

.9992 

.9991 

.9990 

.9990 

.9989 

.9988 

.9987 

0 

0 

0 

0 

0 

3 

.9986 

.9985 

.9984 

.9983 

.9982 

.9981 

.9980 

.9979 

.9978 

.9977 

0 

0 

1 

1 

1 

4 

.9976 

.9974 

.9973 

.9972 

.9971 

.9969 

.9968 

.9966 

.9965 

.9963 

0 

0 

1 

1 

1 

6 

.9962 

.9960 

.9959 

.9957 

.9956 

.9954 

.9952 

.9951 

.9949 

.9947 

0 

1 

1 

1 

1 

6 

.9945 

.9943 

.9942 

.9940 

.9938 

.9936 

.9934 

.9932 

.9930 

.9928 

0 

1 

1 

1 

2 

7 

.9925 

.9923 

.9921 

.9919 

.9917 

.9914 

.9912 

.9910 

.9907 

.9905 

0 

1 

1 

2 

2 

8 

.9903 

.9900 

.9898 

.9895 

.9893 

.9890 

.9888 

.9885 

.9882 

.9880 

0 

1 

1 

2 

2 

9 

.9877 

.9874 

.9871 

.9869 

.9866 

.9863 

.9860 

.9857 

.9854 

.9851 

0 

1 

1 

2 

2 

10 

.9848 

.9845 

.9842 

.9839 

.9836 

.9833 

.9829 

.9826 

.9823 

.9820 

1 

1 

2 

2 

3 

11 

.9816 

.9813 

.9810 

.9806 

.9803 

.9799 

.9796 

.9792 

.9789 

.9785 

1 

1 

2 

2 

3 

12 

.9781 

.9778 

.9774 

.9770 

.9767 

.9763 

.9759 

.9755 

.9751 

.9748 

1 

1 

2 

3 

3 

13 

.9744 

.9740 

.9736 

.9732 

.9728 

.9724 

.9720 

.9715 

.9711 

.9707 

1 

1 

2 

3 

3 

14 

.9703 

.9699 

.9694 

.9690 

.9686 

.9681 

.9677 

.9673 

.9668 

.9664 

1 

1 

2 

3 

4 

15 

.9659 

.9655 

.9650 

.9646 

.9641 

.9636 

.9632 

.9627 

.9622 

.9617 

1 

2 

2 

3 

4 

16 

.9613 

.9608 

.9603 

.9598 

.9593 

.9588 

.9583 

.9578 

.9573 

.9568 

1 

2 

2 

3 

4 

17 

.9563 

.9558 

.9553 

.9548 

.9542 

.9537 

.9532 

.9527 

.9521 

.9516 

1 

2 

3 

4 

4 

18 

.9511 

.9505 

.9500 

.9494 

.9489 

.9483 

.9478 

.9472 

.9466 

.9461 

1 

2 

3 

4 

5 

19 

.9455 

.9449 

.9444 

.9438 

.9432 

.9426 

.9421 

.9415 

.9409 

.9403 

1 

2 

3 

4 

5 

20 

.9397 

.9391 

.9385 

.9379 

.9373 

.9367 

.9361 

.9354 

.9348 

.9342 

1 

2 

3 

4 

5 

21 

.9336 

.9330 

.9323 

.9317 

.9311 

.9304 

.9298 

.9291 

.9285 

.9278 

1 

2 

3 

4 

5 

22 

.9272 

.9265 

.9259 

.9252 

.9245 

.9239 

.9232 

.9225 

.9219 

.9212 

1 

2 

3 

4 

6 

23 

.9205 

.9198 

.9191 

.9184 

.9178 

.9171 

.9164 

.9157 

.9150 

.9143 

1 

2 

3 

5 

6 

24 

.9135 

.9128 

.9121 

.9114 

.9107 

.9100 

.9092 

.9085 

.9078 

.9070 

1 

2 

4 

5 

6 

25 

.9063 

.9056 

.9048 

.9041 

.9033 

.9026 

.9018 

.9011 

.9003 

.8996 

1 

3 

4 

5 

6 

26 

.8988 

.8980 

.8973 

.8965 

.8957 

.8949 

.8942 

.8934 

.8926 

.8918 

1 

3 

4 

5 

6 

27 

.8910 

.8902 

.8894 

.8886 

.8878 

.8870 

.8862 

.8854 

.8846 

.8838 

1 

3 

4 

5 

7 

28 

.8829 

.8821 

.8813 

.8805 

.8796 

.8788 

.8780 

.8771 

.8763 

.8755 

1 

3 

4 

6 

7 

29 

.8746 

.8738 

.8729 

.8721 

.8712 

.8704 

.8695 

.8686 

.8678 

.8669 

1 

3 

4 

6 

7 

30 

.8660 

.8652 

.8643 

.8634 

.8625 

.8616 

.8607 

.8599 

.8590 

.8581 

1 

3 

4 

6 

7 

31 

.8572 

.8563 

.8554 

.8545 

.8536 

.8526 

.8517 

.8508 

.8499 

.8490 

2 

3 

5 

6 

8 

32 

.8480 

.8471 

.8462 

.8453 

.8443 

.8434 

.8425 

.8415 

.8406 

.8396 

2 

3 

5 

6 

8 

33 

.8387 

.8377 

.8368 

.8358 

.8348 

.8339 

.8329 

.8320 

.8310 

.8300 

2 

3 

5 

6 

8 . 

34 

.8290 

.8281 

.8271 

.8261 

.8251 

.8241 

.8231 

.8221 

.8211 

.8202 

2 

3 

5 

7 

8 

35 

.8192 

.8181 

.8171 

.8161 

.8151 

.8141 

.8131 

.8121 

.8111 

.8100 

2 

3 

5 

7 

8 

36 

.8090 

.8080 

.8070 

.8059 

.8049 

.8039 

.8028 

.8018 

.8007 

.7997 

2 

3 

5 

7 

9 

37 

.7986 

.7976 

.7965 

.7955 

.7944 

.7934 

.7923 

.7912 

.7902 

.7891 

2 

4 

5 

7 

9 

38 

.7880 

.7869 

.7859 

.7848 

.7837 

.7826 

.7815 

.7804 

.7793 

.7782 

2 

4 

5 

7 

9 

39 

.7771 

.7760 

.7749 

.7738 

.7727 

.7716 

.7705 

.7694 

.7683 

.7672 

2 

4 

6 

7 

9 

40 

.7660 

.7649 

.7638 

.7627 

.7615 

.7604 

.7593 

.7581 

.7570 

.7559 

2 

4 

6 

8 

9 

41 

.7547 

.7536 

.7524 

.7513 

.7501 

.7490 

.7478 

.7466 

.7455 

.7443 

2 

4 

6 

8 

10 

42 

.7431 

.7420 

.7408 

.7396 

.7385 

.7373 

.7361 

.7349 

.7337 

.7325 

2 

4 

6 

8 

10 

43 

.7314 

.7302 

.7290 

.7278 

.7266 

.7254 

.7242 

.7230 

.7218 

.7206 

2 

4 

6 

8 

10 

44 

.7193 

.7181 

.7169 

.7157 

.7145 

.7133 

.7120 

.7108 

.7096 

.7083 

2 

4 

6 

8 

10 


The precise value of all cosines except cos 0° is less than 1, 































§548 


NATURAL COSINES. 45°-90" 


457 


o 

0.0° 

S 

0.2° 

0.3° 

I 

0.5° 

0.6° 

o 

o 


0.9° 

— Differences 


0' 

6' 

12' 

18' 

24' 

30' 

36' 

42' 

48' 

54' 

1' 

2' 

3' 

4' 

5' 

45 

.7071 

.7059 

.7046 

.7034 

.7022 

.7009 

.6997 

.6984 

.6972 

.6959 

2 

4 

6 

8 

10 

46 

.6947 

.6934 

.6921 

.6909 

.6896 

.6884 

.6871 

.6858 

.6845 

.6833 

2 

4 

6 

8 

11 

47 

.6820 

.6807 

.6794 

.6782 

.6769 

.6756 

.6743 

.6730 

.6717 

.6704 

2 

4 

6 

9 

11 

48 

.6691 

.6678 

.6665 

.6652 

.6639 

.6626 

.6613 

.6600 

.6587 

.6574 

2 

4 

7 

9 

11 

49 

.6561 

.6547 

.6534 

.6521 

.6508 

.6494 

.6481 

.6468 

.6455 

.6441 

2 

4 

7 

9 

11 

60 

.6428 

.6414 

.6401 

.6388 

.6374 

.6361 

.6347 

.6334 

.6320 

.6307 

2 

4 

7 

9 

11 

51 

.6293 

.6280 

.6266 

.6252 

.6239 

.6225 

.6211 

.6198 

.6184 

.6170 

2 

5 

7 

9 

11 

52 

.6157 

.6143 

.6129 

.6115 

.6101 

.6088 

.6074 

.6060 

.6046 

.6032 

2 

5 

7 

9 

12 

53 

.6018 

.6004 

.5990 

.5976 

.5962 

.5948 

.5934 

.5920 

.5906 

.5892 

2 

5 

7 

9 

12 

54 

.5878 

.5864 

.5850 

.5835 

.5821 

.5807 

.5793 

.5779 

.5764 

.5750 

2 

5 

7 

9 

12 

65 

.5736 

.5721 

.5707 

.5693 

.5678 

.5664 

.5650 

.5635 

.5621 

.5606 

2 

5 

7 

10 

12 

56 

.5592 

.5577 

.5563 

.5548 

.5534 

.5519 

.5505 

.5490 

.5476 

.5461 

2 

5 

7 

10 

12 

57 

.5446 

.5432 

.5417 

.5402 

.5388 

.5373 

.5358 

.5344 

.5329 

.5314 

2 

5 

7 

10 

12 

58 

.5299 

.5284 

.5270 

.5255 

.5240 

.5225 

.5210 

.5195 

.5180 

.5165 

2 

5 

7 

10 

12 

59 

.5150 

.5135 

.5120 

.5105 

.5090 

.5075 

.5060 

.5045 

.5030 

.5015 

3 

5 

8 

10 

13 

60 

.5000 

.4985 

.4970 

.4955 

.4939 

.4924 

.4909 

.4894 

.4879 

.4863 

3 

5 

8 

10 

13 

61 

.4848 

.4833 

.4818 

.4802 

.4787 

.4772 

.4756 

.4741 

.4726 

.4710 

3 

5 

8 

10 

13 

62 

.4695 

.4679 

.4664 

.4648 

.4633 

.4617 

.4602 

.4586 

.4571 

.4555 

3 

5 

8 

10 

13 

63 

.4540 

.4524 

.4509 

.4493 

.4478 

.4462 

.4446 

.4431 

.4415 

.4399 

3 

5 

8 

10 

13 

64 

.4384 

.4368 

.4352 

.4337 

.4321 

.4305 

.4289 

.4274 

.4258 

.4242 

3 

5 

8 

11 

13 

65 

.4226 

.4210 

.4195 

.4179 

.4163 

.4147 

.4131 

.4115 

.4099 

.4083 

3 

5 

8 

11 

13 

66 

.4067 

.4051 

.4035 

.4019 

.4003 

.3987 

.3971 

.3955 

.3939 

.3923 

3 

5 

8 

11 

13 

67 

.3907 

.3891 

.3875 

.3859 

.3843 

.3827 

.3811 

.3795 

.3778 

.3762 

3 

5 

8 

11 

13 

68 

.3746 

.3730 

.3714 

.3697 

.3681 

.3665 

.3649 

.3633 

.3616 

.3600 

3 

5 

8 

11 

14 

69 

.3584 

.3567 

.3551 

.3535 

.3518 

.3502 

.3486 

.3469 

.3453 

.3437 

3 

5 

8 

11 

14 

70 

.3420 

.3404 

.3387 

.3371 

.3355 

.3338 

.3322 

.3305 

.3289 

.3272 

3 

5 

8 

11 

14 

71 

.3256 

.3239 

.3223 

.3206 

.3190 

.3173 

.3156 

.3140 

.3123 

.3107 

3 

6 

8 

11 

14 

72 

.3090 

.3074 

.3057 

.3040 

.3024 

.3007 

.2990 

.2974 

.2957 

.2940 

3 

6 

8 

11 

14 

73 

.2924 

.2907 

.2890 

.2874 

.2857 

.2840 

.2823 

.2807 

.2790 

.2773 

3 

6 

8 

11 

14 

74 

.2756 

.2740 

.2723 

.2706 

.2689 

.2672 

.2656 

.2639 

.2622 

.2605 

3 

6 

8 

11 

14 

75 

.2588 

.2571 

.2554 

.2538 

.2521 

.2504 

.2487 

.2470 

.2453 

.2436 

3 

6 

8 

11 

14 

76 

.2419 

.2402 

.2385 

.2368 

.2351 

.2334 

.2317 

.2300 

.2284 

.2267 

3 

6 

8 

11 

14 

77 

.2250 

.2233 

.2215 

.2198 

.2181 

.2164 

.2147 

.2130 

.2113 

.2096 

3 

6 

9 

11 

14 

78 

.2079 

.2062 

.2045 

.2028 

.2011 

.1994 

.1977 

.1959 

.1942 

.1925 

3 

6 

9 

11 

14 

79 

.1908 

.1891 

.1874 

.1857 

.1840 

.1822 

.1805 

.1788 

.1771 

.1754 

3 

6 

9 

11 

14 

80 

.1736 

.1719 

.1702 

.1685 

.1668 

.1650 

.1633 

.1616 

.1599 

.1582 

3 

6 

9 

11 

14 

81 

.1564 

.1547 

.1530 

.1513 

.1495 

.1478 

.1461 

.1444 

.1426 

.1409 

3 

6 

9 

12 

14 

82 

.1392 

.1374 

.1357 

.1340 

.1323 

.1305 

.1288 

.1271 

.1253 

.1236 

3 

6 

9 

12 

14 

83 

.1219 

.1201 

.1184 

.1167 

.1149 

.1132 

.1115 

.1097 

.1080 

.1063 

3 

6 

9 

12 

14 

84 

.1045 

.1028 

.1011 

.0993 

.0976 

.0958 

.0941 

.0924 

.0906 

.0889 

3 

6 

9 

12 

14 

85 

.0872 

.0854 

.0837 

.0819 

.0802 

.0785 

.0767 

.0750 

.0732 

.0715 

3 

6 

9 

12 

14 

86 

.0698 

.0680 

.0663 

.0645 

.0628 

.0610 

.0593 

.0576 

.0558 

.0541 

3 

6 

9 

12 

15 

87 

.0523 

.0506 

.0488 

.0471 

.0454 

.0436 

.0419 

.0401 

.0384 

.0366 

3 

6 

9 

12 

15 

88 

.0349 

.0332 

.0314 

.0297 

.0279 

.0262 

.0244 

.0227 

.0209 

.0192 

3 

6 

9 

12 

15 

89 

.0175 

.0157 

.0140 

.0122 

.0105 

.0087 

.0070 

.0052 

.0035 

.0017 

3 

6 

9 

12 1 

15 


1*6 


All the above cosines are less than 1. 




































458 


NATURAL TANGENTS. 0°-45® supplement 


o 

p 

b 

o 

0.1° 

0.2° 

o 

CO 

b 

o 

o 

0.5° 

0.6° 

O 

0 

i 

0.9° 

-h Differences 


0' 

6' 

12' 

18' 

24' 

30' 

36' 

42' 

48' 

54' 

1' 

2' 

3' 

4' 

6' 

0 

0.0000 

.0017 

.0035 

.0052 

.0070 

.0087 

.0105 

.0122 

.0140 

.0157 

3 

6 

9 

12 

15 

1 

0.0175 

.0192 

.0209 

.0227 

.0244 

.0262 

.0279 

.0297 

.0314 

.0332 

3 

6 

9 

12 

15 

2 

0.0349 

.0367 

.0384 

.0402 

.0419 

.0437 

.0454 

.0472 

.0489 

.0507 

3 

6 

9 

12 

15 

3 

0.0524 

.0542 

.0559 

.0577 

.0594 

.0612 

.0629 

.0647 

.0664 

.0682 

3 

6 

9 

12 

15 

4 

0.0699 

.0717 

.0734 

.0752 

.0769 

.0787 

.0805 

.0822 

.0840 

.0857 

3 

6 

9 

12 

15 

5 

0.0875 

.0892 

.0910 

.0928 

.0945 

.0963 

.0981 

.0998 

.1016 

.1033 

3 

6 

9 

12 

15 

6 

0.1051 

.1069 

.1086 

.1104 

.1122 

.1139 

.1157 

.1175 

.1192 

.1210 

3 

6 

9 

12 

15 

7 

0.1228 

.1246 

.1263 

.1281 

.1299 

.1317 

.1334 

.1352 

.1370 

.1388 

3 

6 

9 

12 

15 

8 

0.1405 

.1423 

.1441 

.1459 

.1477 

.1495 

.1512 

.1530 

.1548 

.1566 

3 

6 

9 

12 

15 

9 

0.1584 

.1602 

.1620 

.1638 

.1655 

.1673 

.1691 

.1709 

.1727 

.1745 

3 

6 

9 

12 

15 

10 

0.1763 

.1781 

.1799 

.1817 

.1835 

.1853 

.1871 

.1890 

.1908 

.1926 

3 

6 

9 

12 

15 

11 

0.1944 

.1962 

.1980 

.1998 

.2016 

.2035 

.2053 

.2071 

.2089 

.2107 

3 

6 

9 

12 

15 

12 

0.2126 

.2144 

.2162 

.2180 

.2199 

.2217 

.2235 

.2254 

.2272 

.2290 

3 

6 

9 

12 

15 

13 

0.2309 

.2327 

.2345 

.2364 

.2382 

.2401 

.2419 

.2438 

.2456 

.2475 

3 

6 

9 

12 

15 

14 

0.2493 

.2512 

.2530 

.2549 

.2568 

.2586 

.2605 

.2623 

.2642 

.2661 

3 

6 

9 

12 

16 

15 

0.2679 

.2698 

.2717 

.2736 

.2754 

.2773 

.2792 

.2811 

.2830 

.2849 

3 

6 

9 

13 

16 

16 

0.2867 

.2886 

.2905 

.2924 

.2943 

.2962 

.2981 

.3000 

.3019 

.3038 

3 

6 

9 

13 

16 

17 

0.3057 

.3076 

.3096 

.3115 

.3134 

.3153 

.3172 

.3191 

.3211 

.3230 

3 

6 

10 

13 

16 

18 

0.3249 

.3269 

.3288 

.3307 

.3327 

.3346 

.3365 

.3385 

.3404 

.3424 

3 

6 

10 

13 

16 

19 

0.3443 

.3463 

.3482 

.3502 

.3522 

.3541 

.3561 

.3581 

.3600 

.3620 

3 

7 

10 

13 

16 

20 

0.3640 

.3659 

.3679 

.3699 

.3719 

.3739 

.3759 

.3779 

.3799 

.3819 

3 

7 

10 

13 

17 

21 

0.3839 

.3859 

.3879 

.3899 

.3919 

.3939 

.3959 

.3979 

.4000 

.4020 

3 

7 

10 

13 

17 

22 

0.4040 

.4061 

.4081 

.4101 

.4122 

.4142 

.4163 

.4183 

.4204 

.4224 

3 

7 

10 

14 

17 

23 

0.4245 

.4265 

.4286 

.4307 

.4327 

.4348 

.4369 

.4390 

.4411 

.4431 

3 

7 

10 

14 

17 

24 

0.4452 

.4473 

.4494 

.4515 

.4536 

.4557 

.4578 

.4599 

.4621 

.4642 

4 

7 

11 

14 

18 

25 

0.4663 

.4684 

.4706 

.4727 

.4748 

.4770 

.4791 

.4813 

.4834 

.4856 

4 

7 

11 

14 

18 

26 

0.4877 

.4899 

.4921 

.4942 

.4964 

.4986 

.5008 

.5029 

.5051 

.5073 

4 

7 

11 

15 

18 

27 

0.5095 

.5117 

.5139 

.5161 

.5184 

.5206 

.5228 

.5250 

.5272 

.5295 

4 

7 

11 

15 

18 

28 

0.5317 

.5340 

.5362 

.5384 

.5407 

.5430 

.5452 

.5475 

.5498 

.5520 

4 

8 

11 

15 

19 

29 

0.5543 

.5566 

.5589 

.5612 

.5635 

.5658 

.5681 

.5704 

.5727 

.5750 

4 

8 

12 

15 

19 

30 

0.5774 

.5797 

.5820 

.5844 

.5867 

.5890 

.5914 

.5938 

.5961 

.5985 

4 

8 

12 

16 

20 

31 

0.6009 

.6032 

.6056 

.6080 

.6104 

.6128 

.6152 

.6176 

.6200 

.6224 

4 

8 

12 

16 

20 

32 

0.6249 

.6273 

.6297 

.6322 

.6346 

.6371 

.6395 

.6420 

.6445 

.6469 

4 

8 

12 

16 

20 

33 

0.6494 

.6519 

.6544 

.6569 

.6594 

.6619 

.6644 

.6669 

.6694 

.6720 

4 

8 

13 

17 

21 

34 

0.6745 

.6771 

.6796 

.6822 

.6847 

.6873 

.6899 

.6924 

.6950 

.6976 

4 

9 

13 

17 

21 

35 

0.7002 

.7028 

.7054 

.7080 

.7107 

.7133 

.7159 

.7186 

.7212 

.7239 

4 

9 

13 

18 

22 

36 

0.7265 

.7292 

.7319 

.7346 

.7373 

.7400 

.7427 

.7454 

.7481 

.7508 

5 

9 

14 

18 

23 

37 

0.7536 

.7563 

.7590 

.7618 

.7646 

.7673 

.7701 

.7729 

.7757 

.7785 

5 

9 

14 

18 

23 

38 

0.7813 

.7841 

.7869 

.7898 

.7926 

.7954 

.7983 

.8012 

.8040 

.8069 

5 

9 

14 

19 

24 

39 

0.8098 

.8127 

.8156 

.8185 

.8214 

.8243 

.8273 

.8302 

.8332 

.8361 

5 

10 

15 

20 

24 

40 

0.8391 

.8421 

.8451 

.8481 

.8511 

.8541 

.8571 

.8601 

.8632 

.8662 

5 

10 

15 

20 

25 

41 

0.8693 

.8724 

.8754 

.8785 

.8816 

.8847 

.8878 

.8910 

.8941 

.8972 

5 

10 

16 

21 

26 

42 

0.9004 

.9036 

.9067 

.9099 

.9131 

.9163 

.9195 

.9228 

.9260 

.9293 

5 

11 

16 

21 

27 

43 

0.9325 

.9358 

.9391 

.9424 

.9457 

.9490 

.9523 

.9556 

.9590 

.9623 

6 

11 

17 

22 

28 

44 

0.9657 

.9691 

.9725 

.9759 

.9793 

.9827 

.9861 

.9896 

.9930 

.9965 

6 

11 

17 

23 

29 


All tangents of angles less than 45° are less than 1 























§548 


NATURAL TANGENTS. 45°-90' 


459 


o 

b 

d 

0.1° 


0.3° 

o 

o 

0.6° 

o 

CO 

d 

o 

o 

1 

0.9° 

+ Differences 


0' 

6' 

12' 

00 

to 

b 

CO 

36' 

42' 

48' 

54' 

1' 

2' 

3' 

4' 

5' 

45 

1.0000 

.0035 

.0070 

.0105 

.0141 

.0176 

.0212 

.0247 

.0283 

.0319 

6 

12 

18 

24 

30 

46 

1.0355 

.0392 

.0428 

.0464 

.0501 

.0538 

.0575 

.0612 

.0649 

.0686 

6 

12 

18 

25 

31 

47 

1.0724 

.0761 

.0799 

.0837 

.0875 

.0913 

.0951 

.0990 

.1028 

.1067 

6 

13 

19 

25 

32 

48 

1.1106 

.1145 

.1184 

.1224 

.1263 

.1303 

.1343 

.1383 

.1423 

.1463 

7 

13 

20 

26 

33 

49 

1.1504 

.1544 

.1585 

.1626 

.1667 

.1708 

.1750 

.1792 

.1833 

.1875 

7 

14 

21 

28 

34 

50 

1.1918 

.1960 

.2002 

.2045 

.2088 

.2131 

.2174 

.2218 

.2261 

.2305 

7 

14 

22 

29 

36 

51 

1.2349 

.2393 

.2437 

.2482 

.2527 

.2572 

.2617 

.2662 

.2708 

.2753 

8 

15 

23 

30 

38 

52 

1.2799 

.2846 

.2892 

.2938 

.2985 

.3032 

.3079 

.3127 

.3175 

.3222 

8 

16 

24 

31 

39 

53 

1.3270 

.3319 

.3367 

.3416 

.3465 

.3514 

.3564 

.3613 

.3663 

.3713 

8 

16 

25 

33 

41 

54 

1.3764 

.3814 

.3865 

.3916 

.3968 

.4019 

.4071 

.4124 

.4176 

.4229 

9 

17 

26 

34 

43 

55 

1.4281 

.4335 

.4388 

.4442 

.4496 

.4550 

.4605 

.4659 

.4715 

.4770 

9 

18 

27 

36 

45 

56 

1.4826 

.4882 

.4938 

.4994 

.5051 

.5108 

.5166 

.5224 

.5282 

.5340 

10 

19 

29 

38 

48 

57 

1.5399 

.5458 

.5517 

.5577 

.5637 

.5697 

.5757 

.5818 

.5880 

.5941 

10 

20 

30 

40 

50 

58 

1.6003 

.6066 

.6128 

.6191 

.6255 

.6319 

.6383 

.6447 

.6512 

.6577 

11 

21 

32 

43 

53 

59 

1.6643 

.6709 

.6775 

.6842 

.6909 

.6977 

.7045 

.7113 

.7182 

.7251 

11 

23 

34 

45 

56 

60 

1.7321 

.7391 

.7461 

.7532 

.7603 

.7675 

.7747 

.7820 

.7893 

.7966 

12 

24 

36 

48 

60 

61 

1.8040 

.8115 

.8190 

.8265 

.8341 

.8418 

.8495 

.8572 

.8650 

.8728 

13 

26 

38 

51 

64 

62 

1.8807 

.8887 

.8967 

.9047 

.9128 

.9210 

.9292 

.9375 

.9458 

.9542 

14 

27 

41 

55 

68 

63 

1.9626 

.9711 

.9797 

.9883 

.9970 

.0057 

.0145 

.0233 

.0323 

.0413 

15 

29 

44 

58 

73 

64 

2.0503 

.0594 

.0686 

.0778 

.0872 

.0965 

.1060 

.1155 

.1251 

.1348 

16 

31 

47 

63 

78 

65 

2.1445 

.1543 

.1642 

.1742 

.1842 

.1943 

.2045 

.2148 

.2251 

.2355 

17 

34 

51 

68 

85 

66 

2.2460 

.2566 

.2673 

.2781 

.2889 

.2998 

.3109 

.3220 

.3332 

.3445 

18 

37 

55 

73 

92 

67 

2.3559 

.3673 

.3789 

.3906 

.4023 

.4142 

.4262 

.4383 

.4504 

.4627 

20 

40 

60 

79 

99 

68 

2.4751 

.4876 

.5002 

.5129 

.5257 

.5386 

.5517 

.5649 

.5782 

.5916 

22 

43 

65 

87 

108 

69 

2.6051 

.6187 

.6325 

.6464 

.6605 

.6746 

.6889 

.7034 

.7179 

.7326 

24 

47 

71 

95 

119 

70 

2.7475 

.7625 

.7776 

.7929 

.8083 

.8239 

.8397 

.8556 

.8716 

.8878 

26 

52 

78 

104 

130 

71 

2.9042 

.9208 

.9375 

.9544 

.9714 

.9887 

.0061 

.0237 

.0415 

.0595 

29 

58 

87 

116 

144 

72 

3.0777 

.0961 

.1146 

.1334 

.1524 

.1716 

.1910 

.2106 

.2305 

.2506 

32 

64 

96 

129 

161 

73 

3.2709 

.2914 

.3122 

.3332 

.3544 

.3759 

.3977 

.4197 

.4420 

.4646 

36 

72 

108 

144 

180 

74 

3.4874 

.5105 

.5339 

.5576 

.5816 

.6059 

.6305 

.6554 

.6806 

.7062 

41 

81 

122 

1631204 

75 

3.7321 

.7583 

.7848 

.8118 

.8391 

.8667 

.8947 

.9232 

.9520 

.9812 






76 

4.0108 

.0408 

.0713 

.1022 

.1335 

.1653 

.1976 

.2303 

.2635 

.2972 






77 

4.3315 

.3662 

.4015 

.4373 

.4737 

.5107 

.5483 

.5864 

.6252 

.6646 






78 

4.7046 

.7453 

.7867 

.8288 

.8716 

.9152 

.9594 

.0045 

.0504 

.0970 


Use ordinary 

79 

5.1446 

.1929 

.2422 

.2924 

.3435 

.3955 

.4486 

.5026 

.5578 

.6140 

interpolation. 

80 

5.6713 

.7297 

.7894 

.8502 

.9124 

.9758 

.0405 

.1066 

.1742 

.2432 






81 

6.3138 

.3859 

.4596 

.5350 

.6122 

.6912 

.7720 

.8548 

.9395 

.0264 






82 

7.1154 

.2066 

.3002 

.3962 

.4947 

.5958 

.6996 

.8062 

.9158 

.0285 






83 

8.1443 

.2636 

.3863 

.5126 

.6427 

.7769 

.9152 

.0579 

.2052 

.3572 






84 

9.5144 

.6768 

.8448 

.0187 

.1988 

.3854 

.5789 

.7797 

.9882 

.2048 






85 

11.430 

11.66 

11.91 

12.16 

12.43 

12.71 

13.00 

13.30 

13.62 

13.95 






86 

14.301 

14.67 

15.06 

15.46 

15.89 

16.35 

16.83 

17.34 

17.89 

18.46 






87 

19.081 

19.74 

20.45 

21.20 

22.02 

22.90 

23.86 

24.90 

26.03 

27.27 






88 

28.636 

30.14 

31.82 

33.69 

35.80 

38.19 

40.92 

44.07 

47.74 

52.08 






89 

57.290 

63.66 

71.62 

81.85 

95.49 

114.6 

143.2 

191.0 

286.5 

573.0 





lESH 


Heavy-face type indicates that the integral part is to be increased by 1, 



























460 


NATURAL COTANGENTS. 0°-45° supplement 


o 

o 

O 

d 


0.2° 

0.3° 

0.4° 

0.5° 

0.6° 

3 

0.8° 

0.9° 

— Differences 


0' 

6' 

12' 

18' 

24' 

d 

CO 

36' 

42' 

48' 

54' 

1' 

2' 

3' 1 

4' 

1 

0 

00 

573.0 

286.5 

191.0 

143.2 

114.6 

95.49 

81.85 

71.62 

63.66 






1 

57.290 

52.08 

47.74 

44.07 

40.92 

38.19 

35.80 

33.69 

31.82 

30.14 






2 

28.636 

27.27 

26.03 

24.90 

23.86 

22.90 

22.02 

21.20 

20.45 

19.74 






3 

19.081 

18.46 

17.89 

17.34 

16.83 

16.35 

15.89 

15.46 

15.06 

14.67 






4 

14.301 

13.95 

13.62 

13.30 

13.00 

12.71 

12.43 

12.16 

11.91 

11.66 

u se orainary 
interpolation. 

6 

11.430 

.2048 

.9882 

.7797 

.5789 

.3854 

.1988 

.0187 

.8448 

.6768 






6 

9.5144 

.3572 

.2052 

.0579 

.9152 

.7769 

.6427 

.5126 

.3863 

.2636 






7 

8.1443 

.0285 

.9158 

.8062 

.6996 

.5958 

.4947 

.3962 

.3002 

.2066 






8 

7.1154 

.0264 

.9395 

.8548 

.7720 

.6912 

.6122 

.5350 

.4596 

.3859 






9 

6.3138 

.2432 

.1742 

.1066 

.0405 

.9758 

.9124 

.8502 

.7894 

.7297 






10 

5.6713 

.6140 

.5578 

.5026 

.4486 

.3955 

.3435 

.2924 

.2422 

.1929 






11 

5.1446 

.0970 

.0504 

.0045 

.9594 

.9152 

.8716 

.8288 

.7867 

.7453 






12 

4.7046 

.6646 

.6252 

.5864 

.5483 

.5107 

.4737 

.4373 

.4015 

.3662 






13 

4.3315 

.2972 

.2635 

.2303 

.1976 

.1653 

.1335 

.1022 

.0713 

.0408 






14 

4.0108 

.9812 

.9520 

.9232 

.8947 

.8667 

.8391 

.8118 

.7848 

.7583 






16 

3.7321 

.7062 

.6806 

.6554 

.6305 

.6059 

.5816 

.5576 

.5339 

.5105 

41 

81 

122 

163 

203 

16 

3.4874 

.4646 

.4420 

.4197 

.3977 

.3759 

.3544 

.3332 

.3122 

.2914 

36 

72 

108 

144 

180 

17 

3.2709 

.2506 

.2305 

.2106 

.1910 

.1716 

.1524 

.1334 

.1146 

.0961 

32 

64 

96 

129 

161 

18 

3.0777 

.0595 

.0415 

.0237 

.0061 

.9887 

.9714 

.9544 

.9375 

.9208 

29 

58 

87 

116 

144 

19 

2.9042 

.8878 

.8716 

.8556 

.8397 

.8239 

.8083 

.7929 

.7776 

.7625 

26 

52 

78 

104 

130 

20 

2.7475 

.7326 

.7179 

.7034 

.6889 

.6746 

.6605 

.6464 

.6325 

.6187 

24 

47 

71 

95 

119 

21 

2.6051 

.5916 

.5782 

.5649 

.5517 

.5386 

.5257 

.5129 

.5002 

.4876 

22 

43 

65 

87 

108 

22 

2.4751 

.4627 

.4504 

.4383 

.4262 

.4142 

.4023 

.3906 

.3789 

.3673 

20 

40 

60 

79 

99 

23 

2.3559 

.3445 

.3332 

.3220 

.3109 

.2998 

.2889 

.2781 

.2673 

.2566 

18 

37 

55 

73 

92 

24 

2.2460 

.2355 

.2251 

.2148 

.2045 

.1943 

.1842 

.1742 

.1642 

.1543 

17 

34 

51 

68 

85 

25 

2.1445 

.1348 

.1251 

.1155 

.1060 

.0965 

.0872 

.0778 

.0686 

.0594 

16 

31 

47 

63 

78 

26 

2.0503 

.0413 

.0323 

.0233 

.0145 

.0057 

.9970 

.9883 

.9797 

.9711 

15 

29 

44 

58 

73 

27 

1.9626 

.9542 

.9458 

.9375 

.9292 

.9210 

.9128 

.9047 

.8967 

.8887 

14 

27 

41 

55 

68 

28 

1.8807 

.8728 

.8650 

.8572 

.8495 

.8418 

.8341 

.8265 

.8190 

.8115 

13 

26 

38 

51 

64 

29 

1.8040 

.7966 

.7893 

.7820 

.7747 

.7675 

.7603 

.7532 

.7461 

.7391 

12 

24 

36 

48 

60 

30 

1.7321 

.7251 

.7182 

.7113 

.7045 

.6977 

.6909 

.6842 

.6775 

.6709 

11 

23 

34 

45 

56 

31 

1.6643 

.6577 

.6512 

.6447 

.6383 

.6319 

.6255 

.6191 

.6128 

.6066 

11 

21 

32 

43 

53 

32 

1.6003 

.5941 

.5880 

.5818 

.5757 

.5697 

.5637 

.5577 

.5517 

.5458 

10 

20 

30 

40 

50 

33 

1.5399 

.5340 

.5282 

.5224 

.5166 

.5108 

.5051 

.4994 

.4938 

.4882 

10 

19 

29 

38 

48 

34 

1.4826 

.4770 

.4715 

.4659 

.4605 

.4550 

.4496 

.4442 

.4388 

.4335 

9 

18 

27 

36 

45 

35 

1.4281 

.4229 

.4176 

.4124 

.4071 

.4019 

.3968 

.3916 

.3865 

.3814 

9 

17 

26 

34 

43 

36 

1.3764 

.3713 

.3663 

.3613 

.3564 

.3514 

.3465 

.3416 

.3367 

.3319 

8 

16 

25 

33 

41 

37 

1.3270 

.3222 

.3175 

.3127 

.3079 

.3032 

.2985 

.2938 

.2892 

.2846 

8 

16 

24 

31 

39 

38 

1.2799 

.2753 

.2708 

.2662 

.2617 

.2572 

.2527 

.2482 

.2437 

.2393 

8 

15 

23 

30 

38 

39 

1.2349 

.2305 

.2261 

.2218 

.2174 

.2131 

.2088 

.2045 

.2002 

.1960 

7 

14 

22 

29 

36 

40 

1.1918 

.1875 

.1833 

.1792 

.1750 

.1708 

.1667 

.1626 

.1585 

.1544 

7 

14 

21 

28 

34 

41 

1.1504 

.1463 

.1423 

.1383 

.1343 

.1303 

.1263 

.1224 

.1184 

.1145 

7 

13 

20 

26 

33 

42 

1.1106 

.1067 

.1028 

.0990 

.0951 

.0913 

.0875 

.0837 

.0799 

.0761 

6 

13 

19 

25 

32 

43 

1.0724 

.0686 

.0649 

.0612 

.0575 

.0538 

.0501 

.0464 

.0428 

.0392 

6 

12 

18 

25 

31 

44 

1.0355 

.0319 

.0283 

.0247 

.0212 

.0176 

.0141 

.0105 

.0070 

.0035 

6 

12 

18 

24 

30 


Heavy-face type indicates that the integral part is to be decreased by 1 

























§548 


NATURAL COTANGENTS. 4*5°-90' 


461 


o 


0.1° 

0.2° 

0.3° 

o 

• o 

0.5° 

0.6° 

0.7° 

0.8° 

0.9° 

- 

Differences 


0' 

6' 

12' 

18' 

24' 

30' 

36' 

42' 

48' 

54' 

V 

2 ' 

3' 

4' 

5' 

45 

1.0000 

.9965 

.9930 

.9896 

.9861 

.9827 

.9793 

.9759 

.9725 

.9691 

6 

11 

17 

23 

29 

46 

0.9657 

.9623 

.9590 

.9556 

.9523 

.9490 

.9457 

.9424 

.9391 

.9358 

6 

11 

17 

22 

28 

47 

0.9325 

.9293 

.9260 

.9228 

.9195 

.9163 

.9131 

.9099 

.9067 

.9036 

5 

11 

16 

21 

27 

48 

0.9004 

.8972 

.8941 

.8910 

.8878 

.8847 

.8816 

.8785 

.8754 

.8724 

5 

10 

16 

21 

26 

49 

0.8693 

.8662 

.8632 

.8601 

.8571 

.8541 

.8511 

.8481 

.8451 

.8421 

5 

10 

15 

20 

25 

50 

0.8391 

.8361 

.8332 

.8302 

.8273 

.8243 

.8214 

.8185 

.8156 

.8127 

5 

10 

15 

20 

24 

51 

0.8098 

.8069 

.8040 

.8012 

.7983 

.7954 

.7926 

.7898 

.7869 

.7841 

5 

9 

14 

19 

24 

52 

0.7813 

.7785 

.7757 

.7729 

.7701 

.7673 

.7646 

.7618 

.7590 

.7563 

5 

9 

14 

18 

23 

53 

0.7536 

.7508 

.7481 

.7454 

.7427 

.7400 

.7373 

.7346 

.7319 

.7292 

5 

9 

14 

18 

23 

54 

0.7265 

.7239 

.7212 

.7186 

.7159 

.7133 

.7107 

.7080 

.7054 

.7028 

4 

9 

13 

18 

22 

55 

0.7002 

.6976 

.6950 

.6924 

.6899 

.6873 

.6847 

.6822 

.6796 

.6771 

4 

9 

13 

17 

21 

56 

0.6745 

.6720 

.6694 

.6669 

.6644 

.6619 

.6594 

.6569 

.6544 

.6519 

4 

8 

13 

17 

21 

57 

0.6494 

.6469 

.6445 

.6420 

.6395 

.6371 

.6346 

.6322 

.6297 

.6273 

4 

8 

12 

16 

20 

58 

0.6249 

.6224 

.6200 

.6176 

.6152 

.6128 

.6104 

.6080 

.6056 

.6032 

4 

8 

12 

16 

20 

59 

0.6009 

.5985 

.5961 

.5938 

.5914 

.5890 

.5867 

.5844 

.5820 

.5797 

4 

8 

12 

16 

20 

60 

0.5774 

.5750 

.5727 

.5704 

.5681 

.5658 

.5635 

.5612 

.5589 

.5566 

4 

8 

12 

15 

19 

61 

0.5543 

.5520 

.5498 

.5475 

.5452 

.5430 

.5407 

.5384 

.5362 

.5340 

4 

8 

11 

15 

19 

62 

0.5317 

.5295 

.5272 

.5250 

.5228 

.5206 

.5184 

.5161 

.5139 

.5117 

4 

7 

11 

15 

18 

63 

0.5095 

.5073 

.5051 

.5029 

.5008 

.4986 

.4964 

.4942 

.4921 

.4899 

4 

7 

11 

15 

18 

64 

0.4877 

.4856 

.4834 

.4813 

.4791 

.4770 

.4748 

.4727 

.4706 

.4684 

4 

7 

11 

14 

18 

65 

0.4663 

.4642 

.4621 

.4599 

.4578 

.4557 

.4536 

.4515 

.4494 

.4473 

4 

7 

11 

14 

18 

66 

0.4452 

.4431 

.4411 

.4390 

.4369 

.4348 

.4327 

.4307 

.4286 

.4265 

3 

7 

10 

14 

17 

67 

0.4245 

.4224 

.4204 

.4183 

.4163 

.4142 

.4122 

.4101 

.4081 

.4061 

3 

7 

10 

14 

17 

68 

0.4040 

.4020 

.4000 

.3979 

.3959 

.3939 

.3919 

.3899 

.3879 

.3859 

3 

7 

10 

13 

17 

69 

0.3839 

.3819 

.3799 

.3779 

.3759 

.3739 

.3719 

.3699 

.3679 

.3659 

3 

7 

10 

13 

17 

70 

0.3640 

.3620 

.3600 

.3581 

.3561 

.3541 

.3522 

.3502 

.3482 

.3463 

3 

7 

10 

13 

16 

71 

0.3443 

.3424 

.3404 

.3385 

.3365 

.3346 

.3327 

.3307 

.3288 

.3269 

3 

6 

10 

13 

16 

72 

0.3249 

.3230 

.3211 

.3191 

.3172 

.3153 

.3134 

.3115 

.3096 

.3076 

3 

6 

10 

13 

16 

73 

0.3057 

.3038 

.3019 

.3000 

.2981 

.2962 

.2943 

.2924 

.2905 

.2886 

3 

6 

9 

13 

16 

74 

0.2867 

.2849 

.2830 

.2811 

.2792 

.2773 

.2754 

.2736 

.2717 

.2698 

3 

6 

9 

13 

16 

75 

0.2679 

.2661 

.2642 

.2623 

.2605 

.2586 

.2568 

.2549 

.2530 

.2512 

3 

6 

9 

12 

16 

76 

0.2493 

.2475 

.2456 

.2438 

.2419 

.2401 

.2382 

.2364 

.2345 

.2327 

3 

6 

9 

12 

15 

77 

0.2309 

.2290 

.2272 

.2254 

.2235 

.2217 

.2199 

.2180 

.2162 

.2144 

3 

6 

9 

12 

15 

78 

0.2126 

.2107 

.2089 

.2071 

.2053 

.2035 

.2016 

.1998 

.1980 

.1962 

3 

6 

9 

12 

15 

79 

0.1944 

.1926 

.1908 

.1890 

.1871 

.1853 

.1835 

.1817 

.1799 

.1781 

3 

6 

9 

12 

15 

80 

0.1763 

.1745 

.1727 

.1709 

.1691 

.1673 

.1655 

.1638 

.1620 

.1602 

3 

6 

9 

12 

15 

81 

0.1584 

.1566 

.1548 

.1530 

.1512 

.1495 

.1477 

.1459 

.1441 

.1423 

3 

6 

9 

12 

15 

82 

0.1405 

.1388 

.1370 

.1352 

.1334 

.1317 

.1299 

.1281 

.1263 

.1246 

3 

6 

9 

12 

15 

83 

0.1228 

.1210 

.1192 

.1175 

;il 57 

.1139 

.1122 

.1104 

.1086 

.1069 

3 

6 

9 

12 

15 

84 

0.1051 

.1033 

.1016 

.0998 

.0981 

.0963 

.0945 

.0928 

.0910 

.0892 

3 

6 

9 

12 

15 

85 

0.0875 

.0857 

.0840 

.0822 

.0805 

.0787 

.0769 

.0752 

.0734 

.0717 

3 

6 

9 

12 

15 

86 

0.0699 

.0682 

.0664 

.0647 

.0629 

.0612 

.0594 

.0577 

.0559 

.0542 

3 

6 

9 

12 

15 

87 

0.0524 

.0507 

.0489 

.0472 

.0454 

.0437 

.0419 

.0402 

.0384 

.0367 

3 

6 

9 

12 

15 

88 

0.0349 

.0332 

.0314 

.0297 

.0279 

.0262 

.0244 

.0227 

.0209 

.0192 

3 

6 

9 

12 

15 

89 

0.0175 

.0157 

.0140 

.0122 

.0105 

.0087 

.0070 

.0052 

.0035 

.0017 

3 

6 

9 

12 

15 


All cotangents of angles greater than 45° are less than 1. 





















462 


PRACTICAL MENSURATION supplement 


Powers and Roots 


No. 

Squares 

Cubes 

Square 

Roots 

Cube 

Roots 

No. 

Squares 

Cubes 

Square 

Roots 

Cube 

Roots 

1 

1 

1 

1.000 

1.000 

61 

2 601 

132 651 

7.141 

3.708 

2 

4 

8 

1.414 

1.260 

52 

2 704 

140 608 

7.211 

3.733 

3 

9 

27 

1.732 

1.442 

53 

2 809 

148 877 

7.280 

3.756 

4 

16 

64 

2.000 

1.587 

54 

2 916 

157 464 

7.348 

3.780 

■ 5 

25 

125 

2.236 

1.710 

55 

3 025 

166 375 

7.416 

3.803 

6 

36 

216 

2.449 

1.817 

56 

3 136 

175 616 

7.483 

3.826 

7 

49 

343 

2.646 

1.913 

57 

3 249 

185 193 

7.550 

3.849 

8 

64 

512 

2.828 

2.000 

58 

3 364 

195 112 

7.616 

3.871 

9 

81 

729 

3.000 

2.080 

59 

3 481 

205 379 

7.681 

3.893 

10 

100 

1 000 

3.162 

2.154 

60 

3 600 

216 000 

7.746 

3.915 

11 

121 

1331 

3.317 

2.224 

61 

3 721 

226 981 

7.810 

3.936 

12 

144 

1 728 

3.464 

2.289 

62 

3 844 

238 328 

7.874 

3.958 

13 

169 

2 197 

3.606 

2.351 

63 

3 969 

250 047 

7.937 

3.979 

14 

196 

2 744 

3.742 

2.410 

64 

4 096 

262 144 

8.000 

4.000 

IS 

225 

3 375 

3.873 

2.466 

65 

4 225 

274 625 

8.062 

4.021 

16 

256 

4 096 

4.000 

2.520 

66 

4 356 

287 496 

8.124 

4.041 

17 

289 

4 913 

4.123 

2.571 

67 

4 489 

300 763 

8.185 

4.062 

18 

324 

5 832 

4.243 

2.621 

68 

4 624 

314 432 

8.246 

4.082 

19 

361 

6 859 

4.359 

2.668 

69 

4 761 

328 509 

8.307 

4.102 

20 

400 

8 000 

4.472 

2.714 

70 

4 900 

343 000 

8.367 

4.121 

21 

441 

9 261 

4.583 

2.759 

71 

5 041 

357 911 

8.426 

4.141 

22 

484 

10 648 

4.690 

2.802 

72 

5 184 

373 248 

8.485 

4.160 

23 

529 

12 167 

4.796 

2.844 

73 

5 329 

389 017 

8.544 

4.179 

24 

576 

13 824 

4.899 

2.884 

74 

5 476 

405 224 

8.602 

4.198 

25 

625 

15 625 

5.000 

2.924 

75 

5 625 

421 875 

8.660 

4.217 

26 

676 

17 576 

5.099 

2.962 

76 

5 776 

438 976 

8.718 

4.236 

27 

729 

19 683 

5.196 

3.000 

77 

5 929 

456 533 

8.775 

4.254 

28 

784 

21 952 

5.292 

3.037 

78 

6 084 

474 552 

8.832 

4.273 

29 

841 

24 389 

5.385 

3.072 

79 

6 241 

493 039 

8.888 

4.291 

30 

900 

27 000 

5.477 

3.107 

80 

6 400 

512 000 

8.944 

4.309 

31 

961 

29 791 

5.568 

3.141 

81 

6 561 

531 441 

9.000 

4.327 

32 

1024 

32 768 

5.657 

3.175 

82 

6 724 

551 368 

9.055 

4.344 

33 

1 089 

35 937 

5.745 

3.208 

83 

6 889 

571 787 

9.110 

4.362 

34 

1 156 

39 304 

5.831 

3.240 

84 

7 056 

592 704 

9.165 

4.380 

35 

1225 

42 875 

5.916 

3.271 

85 

7 225 

614 125 

9.220 

4.397 

36 

1 296 

46 656 

6.000 

3.302 

86 

7 396 

636 056 

9.274 

4.414 

37 

1369 

SO 653 

6.083 

3.332 

87 

7 569 

658 503 

9.327 

4.431 

38 

1444 

54 872 

6.164 

3.362 

88 

7 744 

681 472 

9.381 

4.448 

39 

1 521 

59 319 

6.245 

3.391 

89 

7 921 

704 969 

9.434 

4.465 

40 

1 600 

64 000 

6.325 

3.420 

90 

8 100 

729 000 

9.487 

4.481 

41 

1 681 

68 921 

6.403 

3.448 

91 

8 281 

753 571 

9.539 

4.498 

42 

1 764 

74 088 

6.481 

3.476 

92 

8 464 

778 688 

9.592 

4.514 

43 

1 849 

79 507 

6.557 

3.503 

93 

8 649 

804 357 

9.644 

4.531 

44 

1 936 

85 184 

6.633 

3.530 

94 

8 836 

830 584 

9.695 

4.547 

45 

2 025 

91 125 

6.708 

3.557 

95 

9 025 

857 375 

9.747 

4.563 

46 

2 116 

97 336 

6.782 

3.583 

96 

9 216' 

884 736 

9.798 

4.579 

47 

2 209 

103 823 

6.856 

3.609 

97 

9 409 

912 673 

9.849 

4.595 

48 

2 304 

110 592 

6.928 

3.634 

98 

9 604 

941 192 

9.899 

4.610 

49 

2 401 

117 649 

7.000 

3.659 

99 

9 801 

970 299 

9.950 

4.626 

60 

2 500 

125 000 

7.071 

3.684 

100 

10 000 

1 000 000 

10.000 

4.642 
















§548 


PLANE FIGURES 


463 


Exercises. Mensuration of Plane Figures 

1. Construct a triangle with sides of 2.3'V 3.2", and 3.8" 
respectively. Measure any altitude and find the area of the 
triangle. Check the work by using another altitude. 

In measuring line segments use a pair of dividers to transfer the 
lengths to a graduated ruler. 

2. Draw a circle with a radius of 4.8", take a point 8" 
from the center, and from this point draw a tangent. Find 
the length of the tangent by measuring and check it by § 222. 

3. Construct an equilateral triangle which shall have an 
area of 3 sq. in. Check the work by measuring the base and 
the altitude and finding the area from these results. 

4. If a phonograph record 10" in diameter costs $1.25, and 
if the price is based upon the total area of the disk, how 
much should a record 12" in diameter cost ? 

5. Find the area of a parallelogram of which two sides 
are 2" and 3" respectively, and the included angle is 52° 30'. 
Check the result by measuring an altitude. 

6. Two sides of a parallelogram are 10' and 12' respec¬ 
tively and the included angle is 45°. Find the area and the 
length of each diagonal. 

7. If the diagonal of the □ ABCD is 3.2" and ABAC — 32°, 
what is the area of the rectangle? Check the result by 
drawing the figure and measuring the sides. 

8. The diagonals of a rhombus are 50" and 34" respec¬ 
tively. Find the length of each side, the size of each angle, 
and the area. 

In finding the size of one angle, we have tan a; = f f = 1.4706. Looking 
for 1.4706 in the table of tangents, we find on page 459 that 1.4706 lies 
between 1.4659 and 1.4715, the tangents of 55.7° and 55.8° respectively. 
Since accuracy to the nearest 0.1° is sufficient and 1.4706 is nearer 1.4715 
than it is 1.4659, we see that x = 55.8°. 


464 


PRACTICAL MENSURATION supplement 


9. Within a square which is 2" on a side inscribe the 
largest possible equilateral triangle and find its area. 

The vertices of the triangle must lie on the sides of the square. 

10. Express the results of § 234 and § 235 trigonometri¬ 
cally without using 6'. 

By the aid of Simpson*s Rule find the area between the 
curve and the x axis in Exs. 11 and 12: 

11. Ordinates: 14', 16', 17', 15', 13', 12', 14'; common dis¬ 
tance between the successive ordinates, 4'. 

12. Ordinates: 0", 1.3", 2.4", 3.5", 4.6", 5.7", 6.8", 5.07", 
4.06", 3.05", 2.04"; common distance between the successive 
ordinates, 2.4". 

13. Show that the distance m in miles to the horizon 
from a point f feet above the surface of the sea is given 
approximately by the formula m = ^ 

Use § 253 and take the radius of the earth as 4000 mi. 

14. If the top of a ship’s mast is 60'above the sea, how 
far must the ship sail before it disappears below the 
horizon ? 

15. A sign painter who is laying out a clock face wishes 
to show the time as 8 18. Draw a circle with a radius of 
2" and show accurately the position of the hands. 

16. A hexagonal nut ABCDEF is f" on a side. Find the 

distance across the fiats ” (that is, from AB to DE), and 

the '' distance across the corners ” (that is, AD). 

The quoted terms are used by machinists. 

17. The approximate area A of a segment of a circle 
which is cut off by a chord c and which has a height h 
is given by the formula A = ^hc-\- h^/2 c. What is the 
approximate cross-section area of water flowing through 
a horizontal pipe, when c = 26" and h = 12.5" ? 


§§ 549 , 550 


PROJECTION 


465 


549. Drawing of Solid Figures. The student is expected 
to draw the necessary figures with the help of ruler, com¬ 
passes, and protractor. The measurement of line segments 
is allowed in all cases. 

Solids are represented as if projected upon a plane. We 
may conveniently imagine the projections as shadows cast 
by wire models of the solids, as in the following figures: 





We may think of the shadow as cast by the sun’s rays, which for 
practical purposes are considered parallel. If the rays are perpendic¬ 
ular to the plane of projection, the projection is known as orthogonal 
projection^ the word '' orthogonal ” meaning right-angled. 

The first figure above shows the orthogonal projection of a quadri¬ 
lateral, the second that of a tetrahedron, and the third that of a cube. 

550. Oblique Projection. If the rays which cast the shadows 
are oblique to the plane of projection, the figure formed is 
said to be in oblique projection. 

A convenient method of representing figures in oblique 
projection is illustrated by the cube here shown. In this 
case the rear and front faces (AR^'A'and 
DCC'D') are squares, the lines DA, CB, C'B\ 

D'A’ are parallel, AD'C'B' = 45°, and C'B' is 
half of C'D'. All angles in such a projec¬ 
tion are therefore 45°, 90°, or 135°- 

This type of oblique projection, sometimes called 
cabinet projection, will be used in the exercises which follow. The 
student is expected to be able to reproduce the figures and to draw 
others of the same general nature. 

It is sometimes convenient to take Z.DCB as 30° or as 60°, and to take 
B'C' as three fourths of C'D', but this is not often necessary. 


DL 


0 


\d 

B 


z 


C 


B 




























466 


PRACTICAL MENSURATION supplement 


Exercises. Mensuration of Solids 


1. This figure shows the orthogonal projection upon a 
parallel plane of a rhombus forming the base of a given 
right parallelepiped. It therefore 
shows the base in correct pro¬ 
portions. Measure A BAD with 
the protractor and measure the 
side AB. From these measure¬ 
ments compute the length of each B 

diagonal, and check the results by actual measurement. 

If ZRAZ) = 60° and AR = 0.75", we have ABAC = ZCAD = and 
BD=AB = 0.75". Then, since Z AOB = 90°, we have AO cos 30° 

= 0.75" X 0.8660. Hence AC = 1.5 X 0.8660" = 1.299", or 1.3". 



2. In the figure of Ex. 1 if AB = 14.5" and ZlBAD = 56°, 
what are the lengths of AC and BD^l 


The student should reproduce the figure of Ex. 1 for these propor¬ 
tions. In most exercises it will be sufficient to compute dimensions to 
the nearest 0.1" and angles to the nearest 0.1°. 


3. This figure shows the right parallelepiped of Ex. 1 
projected as described in § 550. 

If the lateral edge AA' is 17.8” 
and the base has the dimensions 
given in Ex. 2, what is the area 
of the total surface of the paral¬ 
lelepiped ? 

The student should draw the figure 
for these dimensions, and similarly in 
each of the exercises where new dimen¬ 
sions are given. 



4. In Ex. 3 compute the length of the longest diagonal of 
the parallelepiped; of the shortest diagonal. 

5. In Ex. 3 compute the volume of the parallelepiped. 








§550 


SOLIDS 


467 


6. This figure shows one of the triangular prisms formed 
by passing planes through the diagonals 
of the figure of Ex. 3. Using the measure¬ 
ments given in Exs. 2 and 3, compute the 
area of the total surface of the prism. 

By referring to the figure of Ex. 1 it is seen 
that ZBOC = 90°. In this and all similar exercises, 
any results obtained in a previous exercise may be 
used in the solution. 



7. This figure shows the triangular prism of Ex. 6 cut 
by a plane through BO and a point Q on 
CC\ Given that ZCOQ = 30°, compute b' 
the area of AOCQ. 



Since, as in all figures in these exercises, the 
projection used is that of § 550, the student will 
find that the AOCQ appears in correct proportions. 
Now, using the length of OC found in Ex. 2, we 
have CQ = OC tan COQ. 



8. In Ex. 7 find the areas of ABQO and BCQ, and the 
area of the total surface of the tetrahedron Q-BCO. 

It will be helpful to draw ABQO and BCQ in correct proportions. 
It is not necessary to draw ABCO since this triangle appears in 
correct proportions in the figure drawn for Ex. 2. 

9. In Ex. 7 find the area of the total surface of the 
truncated prism B'C'O'-BQO. 

10. Show that the area of ABCO in Ex. 8 is equal to the 
area of ABQO multiplied by cos COQ, 

First show that O C = OQ cos COQ, and then use this value in finding 
the area of ABCO. 

11. In the figure of Ex. 7 given that OC = 12.4" and 
CQ = 10.4", find the angle which the plane BQO makes with 
the base. 

Since OB is ± to OC, ZCOQ is the plane angle of ZC-BO-Q. 












468 


PRACTICAL MENSURATION supplement 


12. This figure shows the prism A'-ABO formed by 

passing planes through the diagonals of the figure of Ex. 3. 
In this prism a plane is passed through AB ^ n! 

and a point R on 00'. Also, 05 is J_ to AB 
so that /LOSR is the plane angle of the 
dihedral Z.O-BA-R. Given that the prism 
has the same dimensions as in Exs. 2 and 3 
(that is, that A5=14.5", Z5A0 = 28°, and 
AA'= 17.8"), and that /lOSR = 45°, find the 
lengths of OS and OR, 

Draw the AARO in correct proportions and construct OS X to AB, 
as shown in this figure. Then, using the length ^ ^ 

of OA found in Ex. 2, we have 05= OA sin RAO. 

Hence the length of 05 can be found. Then, since 
Z OSR = 45°, we have OR = 0S. 

13. In Ex. 12 find the areas of A A OR, ARR, and BOR. 

Draw the AABR and BOR in correct proportions. 

14. In Ex. 12 find Z OBR and the length of BR ; find 
/lOAR and the length of AR. 

15. In Ex. 13 show that the area of AABO is equal to the 
area of AABR multiplied by cos 05R. 

16. In Ex. 12 find the volume of the tetrahedron R-ABO 
and the volume of the truncated prism A'R'O'-ARR. 

Ex. 5 may be used in finding the volume of the prism A'-ABO. 

17. In Ex. 12, suppose that ZO5R = 30°, and find the 
lengths of 05 and OR. 

In this case 05 and OR are not equal. 

18. Consider Ex. 13 for the case of Ex. 17. 

19. Consider Ex. 14 for the case of Ex. 17. 

20. Consider Ex. 12, letting Z.BAO = I Z OSR = 15°. 

21. Consider Ex. 13 for the case of Ex. 20. 

22. Consider Ex. 14 for the case of Ex. 20. 










§550 SOLIDS 469 

23. This figure shows the orthogonal projection of a 
regular pyramid with a square base ^ 
upon a plane parallel to the base. 

Given that AB = 24", compute the 
lengths of the diagonals AC and BD, 
and the lengths of the Js OX and OY 
upon AB and BC respectively. 

The student should notice that in this 
projection the vertex V of the pyramid coin- A 
cides with O, the center of the base. 

24. This figure shows the regular pyramid of Ex. 23 
projected as described in § 550. 

Given that the altitude of the 
pyramid is 20.8”, compute the 
volume. 

25. In Ex. 24 compute the 
length of a lateral edge BV of 
the pyramid, and the AVBO 
which this edge makes with 
the base. 

Draw in correct proportions the triangle made by passing a plane 
through VO and the vertex B. 

26. This figure shows the pyramid cut 
from the figure of Ex. 24 by passing planes 
through VO and OX, and through VO and 
OY. Compute the length of the lateral 
edge VY and the ZVYO which VY makes 
with the base. 

27. In Ex. 24 compute the area of the 
total surface of the original pyramid, and 

show that the area of the base is equal to the lateral area 
multiplied by cos VYO. 

The edge VY in Ex. 26 is the slant height of the prigiual pyramid. 



V 











470 


PRACTICAL MENSURATION supplement 



28. This figure shows the base of a regular pyramid 
projected orthogonally upon a par¬ 
allel plane. In the figure, AX, BY, 
and CZ are the altitudes from the 
vertices A, B, and C respectively. If 
AB = 10", what is the area of the 
base of the pyramid? 

The special formula of § 544 for the area 
of an equilateral triangle should be used. 

29. This figure shows the pyramid of Ex. 28 projected 
as described in § 550. If the lateral 
edges make an angle of 45° with the 
base (that is, ZORU= 45°), what is 
the altitude of the pyramid ? 

Draw the A BOV in correct proportions. 

Remember that the medians of a triangle 
meet in a point two thirds of the distance 
from the vertex to the opposite side. 



30. In Ex. 29 compute the volume of the pyramid. 

31. In Ex. 29 compute the slant height VZ and the angle 
made by the lateral faces with the base. 

32. This figure shows the pyramid formed 
by passing planes through VO and OC and 
VO and OF in the figure of Ex. 29. Com¬ 
pute the area of the total surface. 

33. In Ex. 29 how far from the vertex V will a plane 
parallel to the base cut off a pyramid of half the volume ? 

Since the volumes of two similar solids are proportional to the 
cubes of any two corresponding lines, what relation exists between 
the altitude of the small pyramid and that of the original pyramid ? 



34. If the pyramid in Ex. 29 weighs 20 oz., how far from 
the vertex V should a plane parallel to the base be passed 
so as to cut off a small pyramid which shall weigh 5 oz. ? 







§550 


SOLIDS 


471 



35. This figure shows the orthogonal projection of a 

frustum of a regular pyramid with D,_ 

a square base upon a plane parallel 
to the bases. In the figure, O'X' 
and O'Y' are J_ to A'B' and B'C' 
respectively. Given that AB = 14" 
and A'B' = T\ find the areas of 
the upper and lower bases of the 
frustum. 

36. This figure shows the frustum of Ex. 35 projected 
as described in § 550. Given 
that the lateral faces make an 
angle of 60° with the base, 
compute the altitude 00' and 
the slant height YY', 

In the figure the section OYY'O' 
appears in correct proportions. 

37. In Ex. 36 compute the volume of the frustum by 
the formula given in §545. 

38. In Ex. 36 find the area of the 
total surface of the frustum. 

Draw the face ABBA' in correct propor¬ 
tions as here shown. 

39. In Ex. 36 compute the length of BB' and the angles 
which the lateral edges make with the 

lower base and with the upper base. 

Draw the section OBB'O' in correct proportions 
as here shown. 



B" 


40. Draw the figure and compute the 
area of the total surface of the solid formed by passing 
planes through 00' and OX and through 00' and OY in 
the figure of Ex. 36. 


















472 


PRACTICAL MENSURATION supplement 


41. This figure shows the orthogonal projection of a 
frustum of a regular hexagonal pyramid upon a plane 
parallel to the bases. Given that 
O'Z' is _L to A'B\ AR = 20", and 

= find the areas of the 
upper and lower bases. 

The student should use the special 
formula of § 544 for the area of a 
regular hexagon. After finding the 
area of the upper base, that of the 
lower base can be found by an easy 
multiplication. 

42. This figure shows the frustum of Ex. 41 projected 
as described in § 550. If the lat¬ 
eral edges make an angle of 45° 
with the lower base, what is the 
altitude of the frustum ? 

43. In Ex. 42 compute the 
volume of the frustum. 

The student should use the formula in § 545 and the results of Ex. 41. 




44. In Ex. 42 compute the slant height of the frustum 

and the angles which the lateral faces make O!_ Y' 

with the upper and lower bases. 

Draw the section OYY'O' in correct proportions _ 

as here shown. ^ Y" 


45. In Ex. 42 find the number of degrees in each angle 
of a lateral face. 

Draw the face ABB'A' in correct proportions. 

46. In Ex. 42 compute the area of the total surface of 
the frustum. 

47. Draw the figure and compute the area of the total 
surface of the solid formed by passing planes through 00' 
and OX and 00' and 00 in the figure of Ex. 42. 

















§550 


SOLIDS 


473 




48. This figure shows the base of a cylinder of revolution 
projected orthogonally upon a parallel 
plane. Given that the diameter of the 
cylinder is 2.8", compute the circumfer¬ 
ence and the area of the base. 

The tangent lines are helpful in drawing the 
projection of the base, as explained in Ex. 49. 

49. This figure shows the cylinder of Ex. 48 projected 
as described in § 544. If the altitude 
of the cylinder is 3.5", what is the 
volume ? 

It should be noticed that the projections of 
the upper and lower bases are ellipses. In draw¬ 
ing the figure for Ex. 49, the student should draw 
these ellipses freehand. 

50. In Ex. 49 compute the area of the 
curve surface; of the total surface. 

51. This figure shows one fourth of the cylinder of 
Ex. 49, formed by passing planes through 
00' and OB and 00' and OC. Compute the 
area of the total surface of this part of the 
cylinder. 

The difficulties of measuring the complete cylinder 
in projection can usually be avoided by taking one 
fourth of the cylinder as here shown. 

52. The locus of 'points at a distance rfro'm 
a given straight line is the curve surface of a cylinder of rev¬ 
olution which has the radius r and the given line as axis, 

53. The locus of points whose distances from a given plane 
and a given line perpendicular to that plane are in a fixed 
ratio is the curve surface of a cone of revolution which has 
the given line for its axis and the point of intersection of 
the line and the plane for its vertex. 


PS 














474 


PRACTICAL MENSURATION supplement 


54. This figure shows the section of the figure of 
made by a plane passed through OB at an angle 
with the base of the cylinder and cutting the 
element CC' at P. Find the area of the sec¬ 
tion OBP. 

Since OB is ± to the section OCC'O', the dihedral 
angle between the plane OBP and the base is meas¬ 
ured by Z COP. In Ex. 9 we proved that, in the figure 
of Ex. 7, ABCO = ABQO ' cos COQ, and hence we 
may legitimately assume that a similar relation holds 
for this figure. Then since ZCOP and the area of BCO are 
the area of OBP can be found. 



known, 


55. In Ex. 54 compute the area of the curve surface BCP 
and of the entire surface of the solid POBC, 

By analogy to the diagonals of a parallelogram, BP bisects the 
cylindric surface which has the radius OC and the height CP. 

56. This figure shows the intersection of a fourth of a 
cylindric surface and an eighth of a sphere whose center O 
is on the axis of the cylindric sur¬ 
face. If the radius of the cylindric 
surface is half the radius of the 
sphere, the height of the cylindric 
surface is what part of the radius of 
the sphere ? 

Since OP = ^ OP', what is the size of 
ZP'OP^. 

57. Assuming the figure of Ex. 56 completed to show 
the entire hemisphere and the cylindric surface, find the 
ratio of the area of the cylindric surface to the area of 
the hemisphere. 

58. In the figure of Ex. 57 given that the radius of the 
sphere is 4.2”, find the area of the zone whose upper 
base is the intersection of the cylindric surface and the 
hemisphere. 











§550 


SOLIDS 


475 



59. This figure shows the intersection of a fourth of a 
cone of revolution whose vertex is at 0 with an eighth 
of a sphere of center O. Given that the 
radius of the sphere is 6.3" and that 
ZQOO' = 50°, find the radius of the 
circular section made by the conic sur¬ 
face on the sphere. 

60. Assuming the figure of Ex. 59 
completed to show the entire hemi- 5“ 
sphere and the cone, find the ratio of the volume of the 
cone to that of the hemisphere. 

61. In Ex. 60 find the volume of the spherical segment 
which lies above the intersection of the conic surface and 
the hemisphere. 

62. The figure below shows one fourth of a cone of rev¬ 
olution. Given that the radius of the base is 3.5" and that 
ZO VC =30°, compute the altitude of the 
cone and the angle which an element makes 
with the base. 

63. In Ex. 62 compute the volume of 
the entire cone and the area of the curve 
surface. 

64. In Ex. 62 how far from the vertex 
of the cone must the plane B'C'O' be 
passed in order to bisect the volume of the entire cone ? 

65. In Ex. 64 compute the volume of the frustum by the 
formula in § 545 and check by taking half the first result 
of Ex. 63. 

66. Consider Ex. 64, supposing that the plane B'C'O' is 
to cut off a frustum equivalent to one third the cone. 

67. In Ex. 66 compute the volume of the frustum and 
check as in Ex. 65. 








476 


PRACTICAL MENSURATION supplement 


68. This figure shows a cube 18" on an edge projected 
as described in § 550. The midpoint V 
of C'B' is the vertex of an inscribed 
pyramid. Compute the volume and the ^ 
lateral edges of this pyramid. 

Draw the orthogonal projection of the face 
A'B'C'D' as in Ex. 23, and then draw the 
^A'VA in correct proportions. 



69. In Ex. 68 draw each face of the pyramid in correct 
proportions, and compute the altitude of each one. 

70. In Ex. 68 compute the area of the total surface of 
the pyramid. 

71. In Ex. 68 compute the size of each angle of l\VAD\ 
of AUDC; of ARAB. 

72. In Ex. 68 suppose that a plane bisecting the cube is 
passed parallel to the base ABCD, thus cutting off a frustum 
of the pyramid. Draw the figure and compute the volume 
and the area of the total surface of the frustum. 


73. If another pyramid with the base BCC'B and the 
vertex A is inscribed in the cube of Ex. 68, how does its 
volume compare with that of the pyramid V-ABCD ? 


In each case, state the locus of points in space which satisfy 
the following conditions, and draw the figure: 

74. Equidistant from two intersecting planes, and at a 
distance r from a fixed point O. 

75. At a distance d from a line through two fixed points 
A and 0, and at a distance r greater than d from O. 

76. Equidistant from two planes which intersect in the 
line AB and at a distance r from AB. 

77. At a distance D from a given plane m and at a dis¬ 
tance r from a line PQ which is perpendicular to m. 






§550 


SOLIDS 


477 


78. A regular pyramid with a square base 8" on a side 
has an altitude of 10". Compute the volume of the solid 
and the area of the total surface. 

79., Consider Ex. 78 for a regular pentagonal pyramid 
with the same side and altitude. 

80. Consider Ex. 78 for a regular hexagonal pyramid 
with the same side and altitude. 

81. Consider Ex. 78 for a regular octagonal pyramid 
with the same side and altitude. 

82. The lower base of a frustum of a regular pyramid 
is a square 6" on a side, the altitude of the frustum is 8", 
and the side of the upper base is half that of the lower 
base. Compute the volume of the frustum. 

83. In Ex. 82 compute the area of the total surface. 

84. Consider Exs. 82 and 83 for a frustum of a regular 
pentagonal pyramid with the same dimensions. 

85. Consider Exs. 82 and 83 for a frustum of a regular 
hexagonal pyramid with the same dimensions. 

86. Consider Exs. 82 and 83 for a frustum of a regular 
octagonal pyramid with the same dimensions. 

87. A water tank open at the top is to be built of sheet 
iron in the form of a frustum of a cone. The diameter of 
the lower base is to be 14', that of the upper base 10|', and 
the tank is to contain 1500 cu. ft. If an additional 15% 
of the surface area is allowed for waste and overlapping, 
how many square feet of sheet iron are required to build 
the tank ? 

88. A steel container is in the form of a cylinder with a 
hemispherical top. If the inside length of the cylinder is 
4' and the diameter is 1'6", what is the volume ? 

89. In Ex. 88 find the area of the entire inside surface. 


478 


PRACTICAL MENSURATION supplement 


90. A cylindric silo 30' in diameter is covered by the roof 
shown in this figure. The lower part of the roof is in the 
form of a frustum of a cone whose upper base 
has a diameter of 15' and whose sloping sides 
make an angle of 45° with the lower base. The 
upper part is a conic surface which makes an angle of 10° 
with the upper base of the frustum. Making an allowance 
of 20% for waste, compute the number of square feet of 
material required to cover this roof. 

91. Find the diameter of a solid sphere formed by melting 
and recasting the metal in two solid spheres of lead which 
are 2" and 3" respectively in diameter. 

92. If a spherical drop of water is vaporized into a spray 
of 1000 equal spheres, but the total volume of water is un¬ 
changed, by what per cent is the total surface increased ? 

93. A cross section of a hollow cast-iron pillar 10' long is 
a triangle whose sides are 6", 8", and 10", and the thickness 
of the metal is 1". If cast iron weighs 450 lb. per cubic foot, 
how much does the pillar weigh ? 

94. A cubic foot of brass is drawn out into a wire 0.1" in 
diameter. Find the length of the wire. 

95. During a rainfall of 0.5", to what depth will a circular 
well 5' 6" in diameter be filled by the drainage from a 
rectangular asphalted court which is 30'6"X26'5"? 

96. If a plane cuts the axis of a right circular cylinder 
which is 14" in diameter at an angle of 45°, what is the 
area of the section thus formed ? 

If A is the area, then A cos 45° is equal to what area (Ex. 54)? 

97. A hole 1" square is cut horizontally and synmietrically 
through a vertical wooden cylinder which is V2 inches in 
diameter. Find the approximate amount of wood removed 
in cutting the hole. 




§550 . 


SOLIDS 


479 


98. The radii of the bases of a frustum of a right cir¬ 
cular cone are 11" and 12.5" respectively, and the height 
of the,frustum is 5". In the original cone, find the area of 
the curve surface, the area of the total surface, the altitude, 
and the volume. 

99. The area of the total surface of a polyhedron weigh¬ 
ing 64 lb. is 340 sq. in. Find the surface of a similar poly¬ 
hedron made of the same material and weighing 1000 lb. 

100. The bottom and top diameters of a tub are to be 24" 
and 30" respectively. What depth should be allowed so that 
the tub shall hold just 30 gal. (1 gal. = 231 cu. in.)? 

101. A pump has a cylindric barrel 4" in diameter, and 
the volume of water in 1' of the length is pumped at each 
stroke. Find the number of strokes necessary to fill a tub 
in the form of a frustum of a cone 3' in diameter at the 
bottom, 4' in diameter at the top, and 1'6" high. 

102. It is desired to double the capacity of a cylindric 
boiler, but to keep the diameter and height in the same 
ratio. By what per cent will the total area be increased ? 

103. If the area of the surface of a spherical balloon is 
doubled, by what per cent is the circumference increased? 
the diameter ? the volume ? the radius ? 

104. Find the least amount of wood which it is necessary 
to waste in cutting a cube out of a wooden sphere 4" in 
diameter. 

105. Find the volume of the largest sphere that can be 
cut from a cone of revolution 14" high and 12" in diameter. 

106. The specifications for a brass sphere 1" in diameter 
require that the alloy contain one part of zinc to two parts 
of copper by volume. Given that 1 cu. ft. of copper weighs 
550 lb. and that 1 cu. ft. of zinc weighs 428 lb., find the 
weight of the sphere. 


480 


PRACTICAL MENSURATION supplement 


107. The wooden part of a top consists of a conic frus¬ 
tum with a hemispherical end. The greatest and least 
diameters of the frustum are 3.5” and 0.5”, and the slant 
height is 3.5”. Find the total volume. 

108. If an iron sphere 4” in diameter is placed in a 
conic vessel which is full of water and whose altitude and 
diameter are each 5”, how much water will run over ? 

109. Find the volume of the largest sphere that can be 
cut from a metal cone whose base has a diameter of 7” 
and whose slant height is 7”. 

110. If a circular hole 1” in diameter is bored through 
a sphere 2” in diameter and the axis of the hole passes 
through the center of the sphere, what is the volume of 
the part of the sphere that is left ? 

111. A cylinder which is 24” in diameter and 16” in 
height is inscribed in a sphere. Find the area and the 
volume of the sphere. 

112. Prove that the volume of a regular octahedron is 
0.47 approximately, and that the area of the total sur¬ 
face is 3.46 approximately, where s is an edge of the 
solid. From these results find the area and the volume of 
a regular octahedron 3” on an edge. 

113. A stone post is to be surmounted by a sphere 14” 
in diameter, and in order to give the sphere a base upon 
which to stand a segment of one base 2” high is cut from 
the sphere. Find the volume of the segment of the sphere 
that is placed on the post. 

114. If the exposed surface of the segment of the sphere 
in Ex. 113 is polished, how many square inches are polished ? 

115. Find the volume of a spherical sector of a sphere 
of radius r, given that the area of the zone which forms 
the base of the spherical sector is 4 vrrl Draw the figure. 


§551 


FALLACIES 


481 


V. Recreations 

551. Fallacies. Below are given a few curious problems 
and interesting fallacies, generally based upon incorrect 
constructions or statements, which should be undertaken, 
if time allows, simply as recreations. 

1. Any point on a line bisects it. 

In the figure below let BC he any line and P any 

Construct an isosceles AABC upon BC as base, 

and draw AP. 

Since AB = ZCy 

AB = AC, 

and AP = AP, 

then AABP is congruent to AACP. 

Hence BP=PC, B- - 

or any point on a line bisects it. 

2. Every triangle is isosceles. 

Let ABC be any A in which AC is not equal to BC. 

Bisect Z C and construct the ± bisector of AB, letting it meet the 
bisector of ZC at P. They must meet, for if they were II, the bisector 
of Z C would be _L to AB and hence would bisect 
it, thus coinciding with the ± bisector MP. This 
would be possible only if AC = BC, which is con¬ 
trary to what is assumed above. 

Draw PD JL to AC and PE ± to BC. 

Then, since CP bisects ZC, we have PD = PE', 
and since MP is the _L bisector of AB, then AP=BP. 

Then AAPD is congruent to ABPE, and hence AD = BE. 

Similarly, APDC is congruent to APEC, and hence DC = EC. 

Adding, AD + DC = BE + EC, or AC = BC. 

Hence every A is isosceles. 

3. Find the area of this triangle to the nearest 0.1 sq. ft. 

You may use the formula in Ex. 1, page 194, even 
though you have not proved it. If you prefer, draw 
the figure to scale, measure the altitude, and then 
apply § 244. 




point on it. 


A 


148 ' 






482 


RECREATIONS 


SUPPLEMENT 


4. li A>B and B>C, then it follows that A = B=C. 

In this figure the arcs of the (D are all tangent to PT at P. Then, if 
A=ZXPT,B=ZYPT, and C = ZZPT, 

A>B> a 

Now the Z between two (D is defined 
as the Z between their tangents at a 
common point (see page 261). 

But the Z between the tangents at P of any two of these © is 0, 
and hence 

5. Construct a triangle such that the sum of the interior 
angles is less than 180°. 

The three © of which the arcs are here shown 
are tangent at A, B, C. 

Then, as in Ex. 4, the Z between the tangents 
at a common point of any two © is 0. 

Hence the sum of the A of the A formed by the 
tangents at A, B, C is 0. 

6. All circles, however large, have equal circumferences. 

Let two © of unequal radii 

AP and AQ be fastened to¬ 
gether, and let them roll 
along from A to A'. 

Then P reaches P' when Q 
reaches Q'. 

Since the © have rolled along equal distances, their circumferences 
must be equal. 

7. Two coins A and B of the same size are placed upon 
a table so that A is tangent to B. If B is kept fixed and A 
is rolled around R, always remaining tangent to R, how 
many revolutions does A make in rolling once around B ? 

Play fairly; give your answer and reason for it before experimenting. 

8. A man who had a window 2 ft. wide and 4 ft. high 
wished to double its area. He did so, and still the window 
was only 2 ft. wide and 4 ft. high. How was this possible ? 










§551 


FALLACIES 


483 


9. The sum of the parallel sides of a trapezoid is zero. 

In the figure below let ABCD be the trapezoid with bases AB (or h) 
and CD (or a). 

Now let DC be produced to and BA to P so that CS=b and AP = a. 


Then 

and 


APAQ is similar to ASCQ, 
ACDR is similar to AABR. 

z 


Hence 

a _ X 
h y + z X- 

Then * = * 

9 

2 : x + y 

, x — z z 

whence -=- 


z — x x-\-y 

^ , x—z ^ 

But -= —1. 

P^ 

z — x 

Then, by substitution. 


or, multiplying by 6, 

a=— 6; 

whence we have 

(X -j- 6 = 0. 



10. Any number, however large, is equal to zero. 

On a piece of squared paper mark out a square which shall be 8 
by 8, and then draw lines dividing it into three parts A, B, C, as shown. 

Then mark out 
a □ which shall 
be 5 by 13, and di¬ 
vide it into three 
parts such that 
A'= A, B'= B, and 
C' = C, as shown. 

The number of 





































✓ 























A 




-in 



























B 




























































A 


































c 





















B 

9 

































































small squares in the large square is 8 x 8, or 64, and the number of 
small squares in the □ is 5 X 13, or 65. 

Hence 65 = 64, 

or 1 = 0. 

Multiplying these equals by any number, say 25, we have 

25 = 0, 

any number is equal to zero. 


and hence 






















































484 


RECREATIONS 


SUPPLEMENT 


11. From any point outside a line two perpendiculars can 
be constructed to the line. 


Let AB be any line and P any point not on 
AB, and draw PA, PB. I 

With PA and PB as diameters construct © A^ 
intersecting AB at Y and X respectively, and 
draw PX, PE. 


- 


; 


xyx 


Then 

and 

Hence 


ZPXA = 90y 
ZBYP= 90°. 

both PX and PY are ± to AB. 




§173 

§173 


12. The whole of a line is equal to one of its parts. 

In this A, CP is _L to AB, and CX 
is drawn so as to make ZACX=ZB. 

Then AAXC is similar to AACB. 

Hence these A are proportional to 
the squares of corresponding sides, and, 
since they have equal altitudes, to their 

bases also. _ 

AACB B^ 

cx‘" 



X P 


Then 


and hence 


AAXC 

mj' 


AB 

ax' 


Then 


AB 

AC'' + AB^-2AB-AP 


AX ‘ 

ac^'yax'^- 


250, 246 


198,4 


2AX-AP 


or 


whence 


AC^ 

AB 


AB 

+AB- 

A^ 

AB 


AX 


234 


AC^ 




2AP; 


AX= 


AC‘ 


AB, 


or 


AX 

AC^'-AB^AX AC^'-AB^AX 


Hence 


§ 198,2 


AB AX 

AB = AX, 

or the whole of a line is equal to one of its parts. 

13. Show how to arrange six matches so that each match 
shall touch four others. 










§552 


ANCIENT GEOMETRY 


485 


VI. History of Geometry 

552. Ancient Geometry. The geometry of very ancient 
peoples was largely the mensuration of simple areas and 
volumes such as is taught to children in elementary arith¬ 
metic today. They learned how to find the area of a rec¬ 
tangle, and in the oldest mathematical records that we 
have there is some discussion of triangles and of the 
volumes of solids. 

Our earliest documents relating to geometry have come 
to us from Babylon and Egypt. Those from Babylon were 
written, about 2000 B.C., on small clay tablets (some of them 
about the size of the hand) which were afterwards baked 
in the sun. They show that the Babylonians of that period 
knew something of land measures and perhaps had advanced 
far enough to compute the area of a trapezoid. For the 
mensuration of the circle they later used, as did the early 
Hebrews, the value tt = 3. 

The first definite knowledge that we have of Egyptian 
mathematics comes to us from two manuscripts copied on 
papyrus, a kind of paper used in the countries about the 
Mediterranean in early times. One of these manuscripts 
was made by one Aah-mesu (the Moon-born), commonly 
called Ahmes, who fiourished probably about 1550 B.c. The 
original from which he copied, written about 2000 B.c., has 
been lost, but the papyrus of Ahmes, written over three 
thousand years ago, is still preserved and is now in the 
British Museum. In this manuscript, which is devoted 
chiefly to fractions and to a crude algebra, is found some 
work on mensuration. While there is some doubt as to 
the translation of some of the statements, apparently the 
curious rules given include the ones that the area of an 
isosceles triangle is half the product of the base and one of 


486 


HISTORY OF GEOMETRY supplement 


the equal sides, and that the area of a trapezoid with bases 
b, b’ and nonparallel sides each equal to a is ia(b-\-b'). 
One noteworthy advance appears, however, where Ahmes 
gives a rule for finding the area of a circle, substantially 
as follows: Multiply the square on the radius by 



Part of the Ahmes Papyrus 

The oldest extensive book on mathematics in the world, a papyrus roll 
written by Ahmes about 1550 B.C. 

This is equivalent to taking for tt the value 3.1605, and is 
the earliest known case of so close an approximation. 

The second ancient Egyptian manuscript, which may 
have antedated slightly the work of Ahmes, is now in 
Russia. It is on mensuration and apparently contains one 
interesting case of the mensuration of a solid. 































































































§§ 553 , 554 


GREEK GEOMETRY 


487 


553. Early Greek Geometry. From Egypt, and possibly 
from Babylon, geometry passed to the shores of Asia Minor 
and Greece. The scientific study of the subject begins with 
Thales, one of the Seven Wise Men of the early Greek 
civilization. Born at Miletus about 624 B.C., he died there 
about 548 B.c. He founded at Miletus a school of mathe¬ 
matics and philosophy, known as the Ionic School. How 
elementary the knowledge of geometry was at that time 
may be understood from the fact that tra¬ 
dition attributes to Thales only about four 
propositions. 

The greatest pupil of Thales, and one of 
the most remarkable men of antiquity, was 
Pythagoras, born probably on the island 
of Samos, just off the coast of Asia Minor, 
about the year 580 B.c. Pythagoras set 
forth as a young man to travel. He went 
to Miletus and studied under Thales, prob¬ 
ably spent several years in Egypt, and very 
likely went to Babylon. He then founded a school at 
Crotona, in Italy. He is said to have been the first to 
demonstrate the proposition in geometry that the square 
on the hypotenuse of a right triangle is equivalent to the 
sum of the squares on the other two sides. 

554. Euclid. The first great textbook on geometry, and 
the most famous one that has ever appeared, was written 
by Euclid, who taught mathematics in the great university 
at Alexandria, Egypt, about 300 B.C. Alexandria, named in 
honor of Alexander the Great, was then practically a Greek 
city, as it was ruled by the Greeks. 

Euclid's work is known as the Elements, and, in com¬ 
mon with all ancient works, the leading divisions were 
called "'books," as is seen in the Bible and in the works 



Pythagoras 

A coin of Samos, one 
of the oldest known 
portrait medals of a 
mathematician 


488 


HISTORY OF GEOMETRY 


SUPPLEMENT 


of such Latin writers as Csesar and Vergil. This is why 
we speak today of the various books of geometry. In 
this work Euclid placed all the leading propositions of 
plane geometry that were then known, and arranged them 




‘Ifbudanirunaa liber donaitoiom fEoctidis perlpi/ 
CddT^tn arttm/beomorie indpit qpifodiciliiM: 

Qnccuo ett ruiue ps no dl.G inra dl 
logirado fine lantudmccui'’qutcic( 7 / 
trcnittateo ft duo pucta. G jlinca rccca 
i ab vno puctoadaliii b:eutl1ima qnc/ 
lio i crtranitaKs foae vcniq.ico:^ red 
picn0.G0upfidea c q tosim'diiie r lati 
nidine nh bycui^termi qiiide fu t lincc. 
G0upf>dc6 plana c ab vna Itnca ada/ 
Itaertelio icytrcnntatcafuaercdpira 
Gt^nguluaplanuac Diiaru lincarii al/ 

_rernus ptaau8;qiiap crpafio e lap fup/ 

hcic applicanoq, no oireaa. G jQuado aiit anguium ptinct Due 
linecrecterectiline’angiiluonoiaf. Q fi3nrcaalincafuprcaa 
Oetent onoq} angnli vtrobiqj fucrit cqlceco?: vtcrq j rcct^ait 
ciincaq} llnec lupftasa cnHui?ftat ppcndicuUris vocaf .Gan 
gulue VO qiii recto maio: e obnilne Oicit.Gangul^vo mtnoJ re 
etc acur’appdlaf ■GZemnn‘’c qo vniulcoinfq j nma e.G^'igura 
t q tniiiio vl termia pnncf-.GiOrcul‘’c figura plana vna qocm li/ 
nea ptetat qcircufcrpioa noiaf :m cui'^nicdio pucT’c: a quo'oee 
lince reae ad arcuferctiJ creuiee fibiiuicejlu t equalce. i£t bic 
quidepuct^cctru circnli ot.OtOiametCT circuit c linea recta que 
liip ci'^ enpp trulicne c)Trcimtatclq3 fuas circO ferctic applicana 
circulu i DUO media Dinidit.GScmicircnlua c fignra plana Dia/ 
metro arculi i mcdictate orcufCTenticpttnta.Q 1(i>o:no circn / 
It c ftgura plana recta lina i parte cirrii ferene pteta; Icminrcn / 
lo qnidc ant maio: aut minoj. G TRecnlmce figure lut q recne Iw 
iicio corinent quaru qneda trilatae q tnb‘>ecn£i imeta: quedd 
quadniatereq qtuoirectialineie.qdamlhlaterc qua plunbua 
qtqnaroojrecnalineiacontindit.G ■^-Igurarii trilatcraru;alia 
eft tnanguluabneiria latcra cqnaha.3Ua mangulus duo brio 
eqlialatCTa.aiiamanguluomu incqualiom latari. IDapitenl 
alia eft ottbogoniu'.vnu.I rectumanguluin babcno.SItacam' 
bligomnm aliqucm obtulum angnlum habene.3lia eft opgoiii 
nm-.m qua treeangali Innt acun G'Siguraru aute quadrilatcrap 
3 iia eft iidratum quod eft cqmlatcru atq> reaangulu. 3lia el I 
tctTagon^ong’tqtft figura rectangula: led equilataa non eft. 
3Ua eft bdmua^-m; que ell cquilatera: led rectangula non eft. 


JX prinapfis pie nonrepmo ce DlRin^ 
nonibuoearandem. 



Ori|bn>M 

I i^raru* 



First Page of Euclid’s Elements 

From the first printed edition, Venice, 1482 


in a logical order. Most geometries of any importance 
since his time have been based upon this great work of 
Euclid, and improvements in the sequence, symbols, and 
wording have been made as occasion demanded. 






































§§ 555-557 


EUROPEAN GEOMETRY 


489 


555. Geometry in the East. The East did little for geom¬ 
etry, although contributing considerably to algebra. The 
first great Hindu writer was Aryabhatta, who was born in 
47 6 A. D. He gave the very close approximation for tt which 
we express in modern notation as 3.1416. The Arabs, about 
the time of the Arabian Nights tales (800 A.D.), did much 
for mathematics by translating the Greek authors into 
their own language and by bringing learning from India. 
Indeed, it is to the Arab mathematicians of the ninth and 
tenth centuries that modern Europe owes its first knowledge 
of the Elements of Euclid. The Arabs, however, contrib¬ 
uted nothing of importance to geometry. 

556. Geometry in Europe. In the twelfth century Euclid 
was translated from the Arabic into Latin, since Greek 
manuscripts were not then at hand, or were neglected 
because of ignorance of the language. The leading trans¬ 
lators were Adelard of Bath (1120), an English monk who 
had learned Arabic in Spain or in Egypt; Gherardo of 
Cremona, an Italian monk of the twelfth century; and Jo¬ 
hannes Campanus (about 1250), chaplain to Pope Urban IV. 

In the Middle Ages in Europe nothing worthy of note 
was added to the geometry of the Greeks. The first Latin 
edition of Euclid's Elements was printed in 1482, and the 
first English edition in 1570. 

557. Important Propositions. A few facts concerning some 
of the important propositions will be found of interest. 

The theorem which asserts that the base angles of an 
isosceles triangle are equal is said to have been first proved 
by Thales, about 575 B.C. This theorem represented the 
usual limit of instruction in geometry in the Middle Ages, 
and probably on this account was called the pons asino- 
Tum (the bridge of fools); that is, it formed a kind of 
bridge across which fools could not pass. Roger Bacon, 

PS 


490 


HISTORY OF GEOMETRY 


SUPPLEMENT 


about 1250, called it the fuga miserorum (the flight of the 
miserable ones) because they fled at the sight of it. 

The second of the congruence theorems is also attributed 
to Thales, who is said to have used it in measuring the 
distance from the shore to a ship. 

The proposition which relates to the sum of the angles 
of a triangle is referred to by one of the later Greek writers 
in these words: ''The ancients investigated the theorem 
of the two right angles in each individual species of tri¬ 
angle, — first in the equilateral, again in the isosceles, and 
afterwards in the scalene triangle.’’ It is interesting to 
see that we do not have to take this long method of prov¬ 
ing this simple proposition today. It is said that one of 
the earlier writers, Eudemus, who lived about 335 B.C., 
attributed the theorem to the Pythagoreans. 

Perhaps the earliest records of the Pythagorean Theo¬ 
rem are found in Egyptian and Chinese works which are 
of uncertain dates, but were apparently written before 
1000 B.C., or long before Pythagoras lived. In the Chinese 
work the statement reads: " Square the first side and the 
second side and add them together; then the square root 
is the hypotenuse.” The theorem, however, was not proved 
in either of these works. 

558. The Three Famous Problems. The Greeks very early 
found three problems which they could not solve. The first 
was that of trisecting any given angle,—the trisection 
problem; the second was that of constructing a square 
equivalent to a given circle,—the quadrature problem; 
and the third was that of constructing a cube that should 
have a volume twice that of a given cube,—the duplication 
problem. All three are easily solved if we allow other instru¬ 
ments than the ruler and compasses, but they cannot be 
solved by the use of these two instruments alone. 


§§558,559 DIFFICULTIES OF THE STUDENT 


491 


VII. Suggestions to Teachers 

559. Difficulties of the Student. Among the difficulties and 
failures which are encountered by the student, the follow¬ 
ing demand special attention: 

1. Failure to comprehend the purpose of geometry. At 
the beginning of Book I a special effort should be made to 
have the student appreciate the pleasure of' ^ standing upon 
the vantage ground of truth'' and the meaning of a real, 
deductive, scientific proof of a statement. A reasonable 
number of references to geometric forms found in the 
schoolroom, the use of such genuine applications of geo¬ 
metric forms as are within the ability of the student to 
comprehend, and the transfer of the method of geometric 
reasoning to simple problems of life will be found helpful. 
On the other hand, much time can be wasted by dwelling 
upon forms which, while interesting'pictorially, have no 
significant relation to demonstrative geometry and are of 
no particular value as constructions. 

2. Failure to comprehend the technical language. Stu¬ 
dents are often discouraged because they do not clearly 
see the meaning of such terms as median, isosceles, hexa¬ 
gon, and rhombus. This difficulty is easily removed, when 
it is met, by substituting for the term itself a statement of 
its meaning. In general, the teacher should use as simple 
terms as possible, particularly in the first part of the work. 
On this account it is better to use a familiar term such as 

corresponding" instead of ^'homologous," to speak of 
"what is given" rather than of the "hypothesis," and to 
speak of "what is to be proved" rather than of the "con¬ 
clusion," especially as this last term is applied to a state¬ 
ment at the beginning rather than at the end of a proof. 
Simplicity of language and of symbols is a great asset. 


492 


SUGGESTIONS TO TEACHERS supplement 


3. The idea that geometry is to be memorized. This diffi¬ 
culty can be best overcome by paying particular attention 
to the first few propositions. The teacher should develop 
these propositions carefully by questioning the class before 
the theorems are given, and thus lead the students to feel 
that they are discovering the proofs for themselves. The 
students will then come to prefer independent work and 
will thereafter read the proof in the text as a model of 
style rather than as a necessary aid. A second valuable aid 
is the introduction of a number of simple exercises with 
each of the first few propositions. Several of these may 
be given as sight work, with rapid demonstration by the 
student at the blackboard. In this way it will be found 
unnecessary to resort to such a doubtful device as that of 
changing the letters or figures from those which have 
been carefully worked out for the text. 

4. Failure to follow a proof given at the blackboard. A 
prominent cause for this failure is the habit which stu¬ 
dents often form of reading lines and angles by their 
letters without pointing to the figures at the same time. 
No one can follow with ease a demonstration filled with 
expressions like ""AAOB—APRX.** It is much better to 
say angle m and point to it, and it is still better to point 
to a line segment and say '"this line segment ” instead of 
using letters to represent it. Letters are more helpful in 
written work than in oral explanation. Teachers who rec¬ 
ognize this fact will not be disturbed by such convenient 
lettering as ABC and A'B'C'. 

It is also helpful to a class if the student who is demon¬ 
strating a proposition begins his proof by saying, "'The 
general plan of the proof is • • •'' and states the plan. For 
example, he may say, ^'The general plan of this proof is 
to show that this triangle is congruent to that one.’^ This 


§560 


DIFFICULTIES OF THE STUDENT 


493 


method is beneficial not only to the class but to the one 
who is giving the proof. The statement of the plan of the 
proof has been given in the fundamental propositions in 
Books I and II, and after this the student is supposed to 
state the plan in each case for himself, 

5. Failure to state with precision what is given and what 
is to he proved. Time devoted to this difficulty is well spent 
if it leads the student to acquire the habit of precise state¬ 
ment of these parts of a proof in the first exercises which 
he meets. A considerable part of the student's difficulty 
lies in the failure to acquire this habit, and it must be 
acquired early if at all. 

6. Failure to draw the figure when the statement is read. 
It is always a great aid to draw the figure with reasonable 
neatness, and in as general a form as the circumstances 
require, while the theorem, problem, corollary, or exercise 
is being read. Such a habit tends to make the statement 
clear at once and to emphasize precisely what is given and 
precisely what is required. 

560. Methods of Attack. A great deal has been written 
upon the methods of attacking an exercise or a proposition. 
For a beginner, however, it is desirable to keep prominently 
in mind only two methods: 

1. Analysis. The student should early acquire the habit 
of saying, ''I can prove this if I can prove that; I can prove 
that if I can prove this third fact," and so on until he 
reaches some statement which he can prove. He should 
then reverse his reasoning and give the proof step by step 
in proper geometric form. 

2. Indirect Method. In case the analysis does not lead to 
the proof the student should say, "Suppose that the fact 
I am to prove is not true, what follows?" thus taking the 
indirect method described in § 56. 


494 SUGGESTIONS TO TEACHERS supplement 

561. Great Basal Propositions. The teacher will find it 
helpful to call attention from time to time, and especially 
at the close of each book, to the great basal propositions of 
geometry as emphasized in this text. Geometry is peculiar 
among all branches of mathematics, and indeed among all 
the sciences, in its dependence upon a strong chain of truths 
and upon the deductive reasoning which results therefrom. 
The great basal propositions, the links in this chain, should 
therefore be thoroughly mastered. 

562. Language and Symbolism. Teachers are advised not 
only to use as simple language as possible, as already men¬ 
tioned, but to avoid new and unusual terms, especially those 
which do not have general international sanction. In the 
same way it is desirable to avoid local or personal preju¬ 
dices in favor of symbols which are not generally recog¬ 
nized. Such symbols are easily created, and they have 
some advantage as pieces of shorthand; but it should be 
remembered that students are being prepared to read 
general mathematical works, and that for this purpose 
they can best be helped by using only recognized language 
and symbols. Such symbolism as s, a. s. for 'Hwo sides 
and the included angle soon runs into the ridiculous, since 
such forms are neither necessary nor generally helpful. 

It is well to remember, too, that there is considerable 
advantage in lettering and in reading a figure counter¬ 
clockwise, particularly when a refiex angle is met; but this 
is not an arbitrary rule to be followed in all cases. If two 
figures are symmetrically arranged like the triangles in 
§ 47, it is much clearer to read the corresponding letters 
in the same order, as ABC and A'B'C', even though in one 
case they are read clockwise. In other words, any such 
arbitrary rule should be broken without hesitation if there 
is a decided gain in so doing. 


§§ 561-564 


LANGUAGE AND SYMBOLISM 


495 


In the matter of symbolism the teacher will find it much 
better to use Z A, or simply a, instead of Z2. The use of the 
numeral is unnecessary, and there is always some confusion 
in seeming to give a num^ical value to an angle which prob¬ 
ably has an entirely different value from the one stated. 

In general, the teacher will find it helpful to letter figures 
systematically, as has been done in the text. For example, 
on account of the ease with which corresponding parts may 
be recognized, it is more helpful to letter two congruent 
triangles as ABC and A'B'C’ than as DPX and LSV, par¬ 
ticularly as it is neither necessary nor desirable to use the 
letters in giving oral proofs. 

563. Discussions. One of the most valuable features in 
the solution of a problem is the discussion of special cases. 
This is, in general, left for the teacher to initiate. Nearly 
every problem has some special case of interest which 
often leads to the discovery that the solution is impossible 
under certain circumstances. No text can reasonably be 
expected to discuss all these special cases, but the class 
should be encouraged to discover the most interesting ones. 

In particular, it is highly desirable to generalize each 
figure under discussion by studying the various shapes 
assumed when some point or some line of the figure is 
moved about in a plane. 

564. Role of Postulates. The teacher will recognize that 
it is possible to reduce the number of postulates in any 
school geometry. This, however, is not desirable. The 
question to be answered in this connection is. What is 
best for the student at his stage of mental development ? 
In general, within reason, a statement that seems obvious 
to a student may safely, at first reading, be taken as a 
postulate, but it should be proved when the feeling of the 
need for demonstration arises. 


496 


SUGGESTIONS TO TEACHERS supplement 


565. Solid Geometry. The introduction to solid geometry 
should be made slowly. Since the student has been accus¬ 
tomed to seeing only plane figures, the drawing of a solid 
figure in the fiat is confusing at first. The best way for the 
teacher to anticipate this difficulty is to have a few pieces 
of cardboard, a few knitting needles filed to sharp points, 
a pine board about a foot square, and some small corks. 
The cardboard can be used to illustrate planes, and can be 
arranged to show parallel planes, perpendicular planes, or 
planes intersecting obliquely. The knitting needles may 
be stuck into the board to illustrate lines perpendicular or 
oblique to a plane. If two or more lines are to meet in a 
point, the needles may be held together by sticking them 
into one of the small corks. The figures given in the text 
can also be illustrated in this manner. Such homely appa¬ 
ratus, which costs almost nothing and is put together in 
class, seems much more real and is usually more satisfac¬ 
tory than the models which are sold by dealers. 

To have a model for each proposition, or even to have 
a photograph or a stereoscopic picture, is, however, a poor 
educational policy. The pupil must learn very early to 
visualize a solid from the flat figure in outline, just as a 
builder or a mechanic learns to read his working drawings. 
The drawing of the different projections of a solid as 
required in connection with the work on practical mensu¬ 
ration (§§ 543-550) is particularly helpful in this respect. 

The logical processes used in solid geometry do not differ 
essentially from those used in plane geometry, and for this 
reason the treatment of the subject is less extended than 
that found in Books I-V. For beginners it is the custom 
to increase somewhat the number of assumptions and to 
reduce the number of propositions, as compared with the 
work in plane geometry. 


§§ 565-568 


IMPORTANT FORMULAS 


497 


VIII. Important Formulas 


566. Notation. The following notation is used in the for¬ 
mulas of plane geometry: 


a = apothem 
A = area 

а, b,c= sides of AABC 

б, b' = bases 

C = circumference 
d = diameter 


h = height, altitude 
p = perimeter 
r = radius 

s = semiperimeter of AABC; 

that is, s = ^ (a H- 6 + c) 
7r = 3i V-, 3.14, or 3.1416. 


567. Formulas for Lines. The following are important: 


Right triangle, 


§§218,252 

Circle, 

C=27rr 



C = 7rd 

§307 

Radius of circle. 

r = -^ 

2ir 


Equilateral triangle. 



Side of square. 

b = y/A 


568. Areas of Plane Figures. The following are important: 

Rectangle, 

A=bh 

§241 

Parallelogram, 

A=bh 

§243 

Triangle, 

A = jbh 

§,244 


A = y/s(s — a)(s — 

6) (s — c) p. 194 

Equilateral triangle. 

II 


Trapezoid, 

A = j h{b -\-b') 

§^47 

Regular polygon. 

A=^ap 

§282 

Circle, 

A = ^rC 

§308 


A = 7rr^ 

§309 



498 


IMPORTANT FORMULAS 


SUPPLEMENT 


569. Notation. In addition to that of § 566, the following 
notation is used in the formulas of solid geometry: 


area of base 

e = element, lateral edge 
E= spherical excess 
I = slant height 


L — lateral area 
M = area of midsection 
S = area of curve surface 
V = volume 


570. Areas of Solid Figures. The following are important: 


Prism, L = ep § 376 

Regular pyramid, L=\lp § 401 

Frustum of regular pyramid, L = \l{p-\-p') §402 

Cylinder of revolution, S=2 irrh § 424 

Cone of revolution, S=llC §438 

S = irrl § 439 

Frustum of cone of revolution, s=^l{C+C') § 441 

Sphere, =4 ttt^ § 497 

Spherical polygon, S=i\oE7r7^ §504 

Zone, S=2 irrh § 508 

571. Volumes. The following are important: 

Parallelepiped, V=Bh § 393 

Prism or cylinder, V=Bh §§ 395,425 

Cylinder of revolution, V = § 426 

Pyramid or cone, v=lBh § § 407, 443 

Cone of revolution, V=l Trr^h _ § 444 

Frustum of pyramid or cone, V—\h{B-\-B'-\-^BB') § 408 

Frustum of cone of revolution, F= J tt/i( r^H- r'^+ rr') § 445 

Prismoid, V 4M) § 534 

Sphere, F=i § 513 

V=\ird^ 



INDEX 


PAGE 

Abbreviations.x 

Alternation, proportion by . 159 
Altitude 53, 313, 333, 334, 351, 

359, 363, 448 

Angle.4 

acute.9,297 

at center of regular polygon 227 

central.109,131 

complement of an ... . 9 

dihedral.297 

exterior.37, 39 

face.307 

inscribed ....... 130 

of a lune.404 

measure of an.131 

oblique.9, 297 

obtuse.9, 297 

parts of an.4 

parts of a dihedral .... 297 
parts of a polyhedral . . . 307 

plane.. 297,298 

polyhedral. 307,387 

reentrant.61 

reflex.9, 494 

right.8,297 

size of an.4, 7 

size of a dihedral .... 297 

spherical.380 

straight.8,9, 297 

supplement of an ... . 9 

trihedral.308 

Angles, adjacent .... 9, 297 

alternate.39 

classes of polyhedral . . . 308 
PS 499 


PAGE 

Angles, complementary . . 9, 297 
corresponding . . .21,39,166 

equal.7 

equal polyhedral.308 

exterior.39 

exterior-interior.39 

face.307 

interior.39 

of a polygon.61, 387 

supplementary .... 9,297 
symmetric polyhedral . . 309 
made by a transversal . . 39 

of a triangle.5 

vertical.18,297 

Antecedents.157 

Apothem.227 

Arc.11, 235 

major.11 

measure of an.131 

minor.11 

Area.201 

of a circle . . . . . 250,252 
formulas for . 450, 452,497,498 
of an irregular polygon . . 207 
lateral 313,333, 334, 351,359,363 
of alune ....... 404 

of a parallelogram .... 205 

of a rectangle . . . 201,203 
of a regular polygon . . . 235 
Simpson’s rule for .... 451 

of a sphere. 399,403 

of a spherical polygon . . 407 
of a spherical triangle . . 406 
of a trapezoid.207 














































500 


INDEX 


PAGE 

Area, of a triangle . . 194,206 

unit of.201 

Axioms.13,14 

Axis, of a cone ..359 

of a cylinder.351 

Base, of a cone.359 

of a pyramid.333 

of a spherical sector . . . 448 

of a triangle.5,10 

Bases, of a cylinder .... 351 
of a frustum of a cone . . 363 
of a frustum of a pyramid . 334 
of a parallelogram .... 53 

of a prism.313 

of a spherical segment . . 448 

of a trapezoid.53 

Bisector.7,8 

Cavalieri’s Theorem .... 437 
Center, of a circle . . . . . 11 
of a regular polygon . . . 227 

of a sphere.375 

Centers, line of.124 

Centroid.144 

Chord, of a circle.Ill 

of a sphere.375 

Circle.11,109 

area of a. 250,252 

circumscribed . . . 124,149,227 

great.377 

inscribed .... 124,149, 227 

as a limit.250 

measurement of a . . . . 247 

poles of a.377 

properties of a.109 

similar parts of a ... . 235 

small.377 

Circles, concentric . . . 124,227 

escribed.149 

tangent.116 


PAGE 


Circumcenter . . . 

. 143,149 

Circumference . . . 

. .11,250 

Commensurable magnitudes . 129 

Complement . . . . 

... 9 

Composition, proportion by .159 

Cone. 

... 359 

circular. 

... 359 

circumscribed . . 

... 362 

frustum of a . . . 

. . .363 

inscribed . . . . 

... 362 

as a limit . . . . 

. . .363 

oblique. 

... 360 

of revolution . . . 

... 360 

Cones, similar . . . 

. . .370 

Congruence . . . . 

21, 66, 202 

Consequents . . . . 

... 157 

Constant . 

... 249 

Construction . . . . 

. 11, 27, 67 

Converse. 

... 49 

Corollary. 

... 18 

Corresponding parts . 

21,166,169 

Cube. 

... 320 

Curve . 

... 3 

Cylinder. 

... 351 

circular. 

... 351 

circumscribed . . 

... 354 

inscribed . . . . 

... 354 

as a limit . . . . 

... 355 

oblique. 

... 351 

of revolution . . . 

... 351 

right. 

... 351 

Cylinders, similar . . 

... 358 

Decagon . 

... ‘61 

Definitions. 

... 1,7 

Degree. 

... 4 

spherical . . . . 

. . .404 

Diagonal ... 8, 53, 61, 313, 387 

Diameter, of a circle . 

... 11 

of a sphere . . . . 

... 375 

Difference. 

... 3,4 


PS 




























































INDEX 


501 


PAGE 

Dimensions.325 

Directrix. 351,359 

Discussion of a problem 67, 495 
Distance, between parallel lines 53 
between parallel planes . . 293 
of a point from a line ... 53 
of a point from a plane . . 285 
between two points ... 53 

polar.377 

spherical.377 

Division, external.160 

internal.160 

proportion by.159 

proportional.160 

Dodecahedron .... 347,348 

Edge, of a dihedral angle . . 297 
Edges, lateral .... 313,333 
of a polyhedral angle . . . 307 
of a polyhedron . . 2, 313, 348 

Element. 351,359 

Ellipse.360 

Equivalence .... 202, 313,325 
Euler's Theorem . . . 348,429 

Excenters.149 

Excess, spherical.404 

Exclusion, proof by .... 40 
Extremes.157 

Faces, lateral . . . 313, 333, 334 
of a dihedral angle .... 297 
of a polyhedral angle . . . 307 
of a polyhedron . . 2, 313,348 

Fallacies.481 

Figure.2,5,275 

plane rectilinear.5 

rectilinear.5,21 

Figures, congruent 21,202,313,387 
drawing ... 11, 67,93,465,493 

similar. 166,431 

Foot of a perpendicular . . 8, 279 

PS 


PAGE 

Formulas, of area 450,452,497,498 


trigonometric.453 

of volume. 452,498 

Frustum, of a cone .... 363 

as a limit.363 

of a pyramid.334 

Generatrix. 351,359 

Geometry, history of ... . 485 
nature of.1,275 

Half-planes.297 

Height . 53,313,333,334,351,359 
slant .... 333,334,360,363 

Hexagon.61 

Hexahedron. 347,348 

History of geometry .... 485 
Homologous parts . . . .21,166 

Hyperbola.360 

Hypotenuse .... 10,173,213 
Hypothesis.20 

Icosahedron. 347,348 

Illusions, optical.13 

Incenter.143,149 

Inclination of a line .... 303 
Incommensurable cases . . . 425 
Incommensurable magnitudes 129 

Inequalities.4,81 

Infinity.250 

Instruments.67 

Intercept.109 

Inversion, proportion by . . 159 

Legs of a triangle.10 

Limit.249 

circle as a.250 

cone as a.363 

cylinder as a.355 

frustum of a cone as a . . 363 
sphere as a.411 














































502 


INDEX 


PAGE 


iJlXlt:. 

of centers .... 

... 124 

curve . 

... 3 

geodetic. 

. . .401 

horizontal .... 

... 8 

straight. 

. . 3,11 

vertical. 

... 8 

Lines, concurrent . . 

... 94 

corresponding . . 

21,166,169 

equal. 

... 3 

parallel. 

. .39,293 

perpendicular. . . 

... 8 

proportional . . . 

. 160,166 

skew. 

... 288 

Locus . . . . . . 

139,140, 285 

Lune. 

. 404,407 

Magnitudes . . . . 

. . 2,129 

Mean proportional. . 

. 157,239 

Means. 

... 157 

Measure. 

... 129 

angle. 

. . .131 

' arc .. 

... 131 

common. 

... 129 

numerical .... 

. . .129 

Measurement, of a circle . . 247 

of a sphere.... 

... 399 

Median. 

. . 94,144 

Mensuration, practical 

. . .449 

Midpoint. 

... 7 

Midsection. 


Multiple. 

. . .129 

Nappe. 

... 359 

Oblique, to a line . . . 

. . . 8,9 

to a plane. 


Octagon. 

... 61 

Octahedron .... 

. 347,348 

Originals, attacking . 

93, 94, 493 

prthocenter- .. .. ! . 

. . .144 


PAGE 

Parabola.360 

Parallel, to a line . . . .39,293 

to a plane.279 

Parallelepiped.320 

oblique.320 

rectangular.320 

right.320 

Parallelogram.53 

Pentagon.61 

Pentagram.63 

Perigon.9 

Perimeter.5, 61, 250 

Perpendicular, foot of a. .8, 279 

to a line.'8, 279 

to a plane.279 

Pi (tt). 251, 254 

Plane.2,275 

tangent to a cone .... 362 

tangent to a cylinder . . . 354 

tangent to a sphere . . . 377 

Planes, parallel.279 

perpendicular.298 

Point.2 

of contact.116 

of tangency.116 

Poles of a circle.377 

Polygon.61 

center of a regular.... 227 

circumcenter of a . . . . 149 

circumscribed.124 

concave . ..61,387 

convex.61, 387 

equiangular.61 

equilateral.61 

incenter of a.149 

inscribed.124 

irregular ..207 

radius of a regular .... 227 

regular.61, 227 

spherical .387 

vertices of a.61,387 

PS 
































































INDEX 


503 


PAGE 

Polygons, mutually equiangular 61 


mutually equilateral ... 61 

similar.166 

Polyhedron.313 

circumscribed.381 

inscribed.381 

regular. 347,348 

relation of parts of a . . . 348 

Polyhedrons, similar .... 431 

Postulates . 13,16, 39,109,139, 

249, 276, 495 

Prism.313 


circumscribed.354 

inscribed.354 

oblique.314 

right.314 

right section of a ... . 314 

triangular.314 

truncated.314 

Prismoid.437 

Problem.18,67 

Problems, famous.490 

Projection, cabinet .... 465 
of a line segment .... 173 

oblique.465 

orthogonal.465 

Proof, analytic . . . . .94,493 

by exclusion.40 

indirect.40,493 

methods of . . 19, 20, 40, 93, 

* 94,140, 493 

nature of a.18,20 

synthetic.94 

Proportion.157 

Divine.239 

division of lines in . . 160,239 

laws of ..158 

nature of quantities in a . 160 
Proportional, fourth .. . . . 185 

mean. 157,239 

Proposition . . . . . 18 

PS 


PAGE 

Pyramid.333 

circumscribed.362 

frustum of a.334 

inscribed ..362 

regular ....... 333,334 

Pythagorean Theorem . 173,213 

Quadrant. 130,377 

Quadrilateral.53,61 

Radius, of a circle.11 

of a regular polygon . . . 227 

of a sphere.375 

Ratio. 129, 203 

extreme and mean .... 239 
incommensurable .... 130 

of similitude.166 

Ray.3 

Recreations of geometry . . 481 

Rectangle. 2, 53, 203 

Reductio dd absurdum ... 40 
Revolution, cone of ... . 360 

cylinder of.351 

similar cones of.370 

similar cylinders of . . . 358 

Rhombus.53 

Rigidity, unit of.34 

Roots, table of.462 

Secant.124,175 

Section, conic.360 

Golden.239 

right . ..314,355 

transverse.355 

Sector, of a circle . . . 130,235 

spherical.448 

Segment, of a circle . . 130, 235 

line. 3 

spherical.448 

Semicircle ..11 

Semicircumference .... 11 



















































504 


INDEX 


PAGE 

Sides, adjacent. 5 

of an angle.4 

corresponding . . 21,166,169 

of a polygon.61,387 

of a triangle.6,10 

Similarity . . . 166,358,370,431 
Similitude, ratio of .... 166 

Simpson’s Rule.451 

Solid. 2,275,320 

rectangular.2,320 

Solution, nature of a . . . . 67 

Sphere.375 

area of a. 399,403 

circumscribed.... 375,381 

inscribed.381,411 

as a limit.411 

properties of a.375 

volume of a ... 399, 411, 412 

Spheres, tangent.377 

Square.8 

Subtend.. 109, 111 

Sum.3, 4 

of angles of a triangle . . 46 

Superposition.23 

Supplement.9 

Surface, conic.359 

cylindric.351 

plane.2,275 

spherical.375 

Symbols.x, 494 

Tables, of powers and roots . 462 
trigonometric.... 454-461 
Tangency, point of .... 116 

Tangent, common.116 

to a circle.116 

to a sphere.377 

Terms of a proportion . . . 157 
Tetrahedron . . . .333,347,348 

Theorem.18 

Transversal.39 

PS 


PAGE 

Trapezoid.53 

Triangle.5,61 

acute.10,387 

ambiguous case of a . . . 78 

birectangular.394 

centers of a . . .143,144,149 

equiangular.10 

equilateral.10,387 

isosceles. 10, 387, 396 

obtuse.10,387 

polar.391 

right. 10,387,394 

scalene.10, 387 

spherical. 387 

trirectangular.394 

Triangles, symmetric.... 396 
Trigonometry .... 274,453 
Trisecting a magnitude . . 94, 490 

Unit.129 

of area.201 

of volume.325 

Variable.249 

Vertex 2, 4, 5,10, 61, 307, 313, 

333, 351, 359, 387 

Volume.325 

of a cone.366 

of a cylinder.357 

formulas for .... 452, 498 
of a frustum .... 344, 367 

of a parallelepiped .... 326 

of a prism.330 

of a prismoid .438 

of a pyramid.341 

of a rectangular parallele¬ 
piped . 325,428 

of a sphere .... 399, 411, 412 
unit of.325 

Zone.409 













































































